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In this video, we will learn how to identify features of quadratic functions, such as the vertex, zeros, axis of symmetry, domain, and range.
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Weβll look at how we can determine these features both graphically and from the equation of the function.
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You should already be familiar with the process of completing the square or writing a quadratic in completed square form, although this will be briefly recapped in the context of examples.
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We recall firstly that a quadratic function is of the general form π of π₯ equals ππ₯ squared plus ππ₯ plus π, where π, π, and π are constants and π must be nonzero.
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An alternative form in which quadratic functions can be represented is completed square or vertex form.
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π of π₯ equals π multiplied by π₯ plus π all squared plus π, where π, π, and π are constants and again π must be nonzero.
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If we were to plot a graph of π¦ equals π of π₯ for a quadratic function, then we find that all quadratic functions share the same general shape, which is known as a parabola.
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The first distinction we can make is in the type of parabola, and this is determined by the sign of the coefficient of π₯ squared.
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Thatβs the value of π.
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If π is positive, then the parabola will be curved upwards as in the diagram on the left, whereas if π is negative, the parabola will curve downwards as in the diagram on the right.
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This is the first key thing to look for when determining the shape of the graph of a quadratic function.
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Letβs think about some of the other general features of quadratic functions that we need to be aware of.
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And weβll do this by considering the graph of a simple quadratic.
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π of π₯ equals π₯ squared plus two π₯ minus three.
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The first thing we can determine about this graph is its π¦-intercept.
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We recall that everywhere on the π¦-axis, π₯ is equal to zero.
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So by substituting zero into the equation of π of π₯, we find the value of π¦ when π₯ equals zero is negative three.
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Now, this is the constant term in our quadratic function, and this will always be the case.
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So in general, if we have a quadratic π of π₯ equals ππ₯ squared plus ππ₯ plus π, then the value of its π¦-intercept will be π.
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The second key feature of a quadratic function is its roots or zeroes.
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Now, these are the π₯-values at which the graph crosses the π₯-axis.
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We know that everywhere on the π₯-axis, π¦ or π of π₯ is equal to zero.
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So these are the solutions to the equation π of π₯ equals zero.
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We can find these values by considering the factored or factorized form of our quadratic function.
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In this case, our quadratic can be factored as π₯ plus three multiplied by π₯ minus one.
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And we then take each of these factors in turn, set them equal to zero, and solve the resulting linear equations, giving π₯ equals negative three and π₯ equals one.
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We now have enough information to be able to sketch this quadratic reasonably accurately.
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The coefficient of π₯ squared is one, so itβs positive, meaning the parabola opens upwards.
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We have a π¦-intercept of negative three and π₯-intercepts of negative three and one.
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The next key feature we want to consider is the vertex or turning point of our quadratic.
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Now, this turning point will be a minimum when the value of π is positive, and it will be a maximum when the value of π is negative.
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In our case, itβs a minimum point.
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Itβs the coordinates of this point here.
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In order to find the coordinates of this point, we consider the vertex form of our quadratic, which in this case is π₯ plus one all squared minus four.
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Weβll review how to do this in some examples.
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Now, we need to recall a general result here, which is that for the quadratic in its general vertex form ππ₯ plus π all squared plus π, its vertex will be at the point negative π, π.
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Which means for our quadratic, its vertex will be at the point negative one, negative four.
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Which makes sense when we consider the position of this point relative to the values weβve marked on our axes.
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So thatβs three key features of our quadratic functions.
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Letβs now consider some more.
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A parabola is a smooth, symmetrical curve, which means that every quadratic graph has an axis or line of symmetry.
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This will be a vertical line passing through the vertex of our function.
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Vertical lines have equations of the form π₯ equals constant.
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And the π₯-value through which this line passes is the π₯-coordinate of the vertex.
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So the equation of the axis of symmetry for this quadratic will be π₯ equals negative one.
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The two remaining features we need to consider are the domain and range of our quadratic.
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Now, we recall firstly that the domain of a function is the set of all values on which the function acts, which we can also think of as the input values to the function.
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A quadratic is just a type of polynomial, and all polynomials can act on all π₯-values.
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This means that there are no restrictions on the values of π₯ on which the function can act.
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So we say that the domain is the set of all real numbers.
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Finally, we consider the range of the function, which is the set of all values the function produces.
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Or in the case of a graph, we can think of it as all the values of π of π₯ or π¦.
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From our graph, we can see that all the possible π¦-values of the function are the values of π¦ from the minimum point upwards.
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Thatβs all π¦-values greater than or equal to negative four.
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We can either express the range as π of π₯ is greater than or equal to negative four.
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Or we can write this using interval notation as the interval from negative four to β, which is closed at the lower end and open at the upper end.
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Now that weβve seen how to identify the key features of quadratic functions, letβs consider some examples.
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Find the coordinates of the vertex of the graph π of π₯ equals π₯ squared plus eight π₯ plus seven.
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State the value of the function at the vertex and determine whether it is a minimum or maximum value.
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In order to find the coordinates of the vertex of this graph, we need to convert its equation to vertex form.
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π of π₯ equals π multiplied by π₯ plus π all squared plus π.
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Now, looking at the equation of this graph, we see that the value of π, the coefficient of π₯ squared, is one.
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So weβre actually just looking for this quadratic in the form π₯ plus π all squared plus π, which we can do by completing the square.
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Firstly, we determine the value of π inside the parentheses.
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And this is always half the coefficient of π₯ in the equation.
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Half of eight is four, so we have π₯ plus four all squared.
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Now, we want this first part of our quadratic function to be equivalent to π₯ squared plus eight π₯.
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But we know that if we were to distribute π₯ plus four all squared, it would give π₯ squared plus eight π₯ plus 16.
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So we have an extra 16 which we need to subtract in order to ensure that these two parts of the quadratic are equivalent.
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π₯ squared plus eight π₯ is therefore equivalent to π₯ plus four all squared minus 16.
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And then we also have the positive seven, which remains the same.
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That value of 16 that weβre subtracting is four squared.
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Itβs the square of our value π.
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Then we just need to simplify.
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Negative 16 plus seven is negative nine.
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So we now have our quadratic in its vertex form.
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We then recall that for a quadratic in its vertex form, its vertex will have the coordinates negative π, π.
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For our quadratic, the value of π is four and the value of π is negative nine.
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So the coordinates of the vertex will be negative π, thatβs negative four, π, which is negative nine.
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So weβve found the coordinates of the vertex of this graph.
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The question also asks us to state the value of the function at the vertex.
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The value of the function will be the π¦-coordinate, so the value is negative nine.
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Finally, we were asked to determine whether this is a minimum or maximum value.
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Well, this is determined by the coefficient of π₯ squared, the value of π, which in our equation is one.
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As π is positive, the parabola will open upwards, which means that the vertex will be a minimum.
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So weβve completed the problem.
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The coordinates of the vertex are negative four, negative nine.
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The value of the function itself is negative nine.
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And this is a minimum value.
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In our next example, weβll see how to determine the domain and range of a quadratic function given in its vertex form.
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Determine the domain and the range of the function π of π₯ equals four multiplied by π₯ minus four all squared minus three.
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Firstly, we recall that the domain is the set of all values on which the function acts, which we can also think of as the set of input values to the function.
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As the function π of π₯ is a polynomial and, more specifically, a quadratic, there are no restrictions on what values it can act on.
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Therefore, we say that the domain of this function is the set of all real numbers.
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The range of a function is the set of all values the function produces, which we can think of as the set of all output values.
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To determine the range of a quadratic function, we can consider its turning point.
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Now, this quadratic function has been given to us in its completed square or vertex form.
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π of π₯ equals π multiplied by π₯ plus π all squared plus π.
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And we know that when a quadratic function is given in this form, its vertex has the coordinates negative π, π.
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The value of π for our quadratic is negative four, and the value of π is negative three.
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So the vertex will be at negative negative four, thatβs four, negative three.
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As the value of π, the coefficient of π₯ squared in our quadratic function, is four, which is positive, we know that its graph will be a parabola which curves upwards.
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So this vertex of four, negative three will be a minimum point.
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The possible values of π of π₯ then will be all the values from this minimum value of the function negative three upwards.
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We can express this either as π of π₯ is greater than or equal to negative three or using interval notation as the interval from negative three to β, which is closed at the lower end and open at the upper end.
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We can answer the problem then by saying that the domain of this function is the set of all real numbers and the range is the interval from negative three to β, which is closed at the lower end and open at the upper end.
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In this example, weβll see how we can use key features of a quadratic function to identify its graph.
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For the function π of π₯ equals π₯ squared minus four π₯ plus three, answer the following questions.
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Firstly, find by factoring the zeroes of the function.
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Secondly, identify the graph of π.
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There are also two further parts to this question.
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So firstly, weβre asked to find the zeros of this function.
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And the method weβre told to use is factoring.
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We therefore need to write our quadratic as the product of two linear factors.
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As the coefficient of π₯ squared is one, we know that the first term in each of our parentheses will be π₯.
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Weβre then looking for two numbers whose sum is the coefficient of π₯, thatβs negative four, and whose product is the constant term, thatβs positive three.
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Well, the two numbers that fit both of those criteria are negative one and negative three.
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Negative one plus negative three is negative four, and negative one multiplied by negative three is positive three.
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So our quadratic factors as π₯ minus one multiplied by π₯ minus three, which we can of course confirm by redistributing the parentheses if we wish.
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We need to use this factored form to determine the zeroes of the function, which we recall are the π₯-values such that π of π₯ equals zero.
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If we set this factored form equal to zero, we then recall that for the product of two things to be zero, at least one of them must themselves be zero.
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So we can take each factor in turn and set it equal to zero, giving two simple linear equations.
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The first can be solved by adding one to each side to give π₯ equals one, and the second can be solved by adding three to each side to give π₯ equals three.
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The roots or zeros of this function then are the values one and three.
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Now, in the second part of the question, weβre asked to identify the graph of our function π.
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And we can see that weβve been given three possibilities: a blue one, a red one, and a green one.
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Now, weβve just found that our graph has zeros at one and three.
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And remember, these zeros are the values of π₯ at which the graph crosses the π₯-axis.
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So if our graph crosses the π₯-axis at one and three, we can see from the figure that this leaves just the red and green graphs.
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The blue graph crosses the π₯-axis or has zeros at values of negative one and negative three.
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Now, we just need to decide between the red and green graphs, which we see are mirror images of each other.
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One is an upward-curving parabola, and the other is a downward-curving parabola.
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We recall that the type of parabola we have will be determined by the value of π.
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Thatβs the coefficient of π₯ squared.
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In our function, the coefficient of π₯ squared is one.
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Itβs a positive value, which means the parabola will curve upwards.
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That means then that the graph of our function π must be the red graph.
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It has the correct zeros and the correct shape.
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We can also see that the π¦-intercept of this graph is three, which is indeed the constant term in our function π of π₯.
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The remaining two parts of the question, which I didnβt write down initially because they give the game away for the previous part are.
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Write the equation for π, the function that describes the blue graph.
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And write the equation for β, the function that describes the green graph.
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Letβs look at this blue graph first of all then.
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We already said that it has zeros at negative one and negative three.
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This means that in its factored form, it has factors of π₯ plus one and π₯ plus three.
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But there could also be a factor of π that we multiply by.
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To determine whether the value of π is one or something else, we consider the π¦-intercept of the graph, which we can see is the same as the π¦-intercept of the red graph.
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Itβs three.
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When we multiply these two factors together, the constant term will be one multiplied by three, which is indeed three.
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And so this tells us that the value of π is simply one.
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Our function π in its factored form then is π₯ plus one multiplied by π₯ plus three.
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If we distribute the parentheses, we have π of π₯ equals π₯ squared plus four π₯ plus three.
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For the green graph, it has the same zeros as our function π.
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So it can be written as π multiplied by π₯ minus one multiplied by π₯ minus three.
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And again, we need to determine whether the value of π is one or something else.
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Well, the π¦-intercept for the green graph is negative three.
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If we multiply together negative one and negative three, we get a value of positive three.
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And so in order to ensure the π¦-intercept, the constant term in the expanded form of β of π₯, is negative three, we need the value of π to be negative one.
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The equation β of π₯ then is negative π₯ minus one multiplied by π₯ minus three.
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In fact, it is the complete negative of our function π of π₯, which we can also see because they are reflections of one another in the π₯-axis.
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We can write the equation β of π₯ then as the complete negative of our function π of π₯.
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β of π₯ is equal to negative π₯ squared minus four π₯ plus three.
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Letβs now summarize some of the key points from this video.
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Quadratic functions can be expressed in their expanded form, π of π₯ equals ππ₯ squared plus ππ₯ plus π.
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Or their completed square or vertex form, π of π₯ equals π multiplied by π₯ plus π all squared plus π, where π, π, π, π, and π are all constants and π must be not equal to zero.
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The graph of a quadratic function is a parabola.
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And if π is positive, the parabola will open upwards, whereas if π is negative, the parabola will open downwards.
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The turning point or vertex of a quadratic can be found from its completed square or vertex form.
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And in the general case, the vertex will have coordinates negative π, π.
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The parabola will also have an axis of symmetry, which is a vertical line passing through this point, with the equation π₯ equals negative π.
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The domain of any quadratic function is the set of all real numbers, unless specified otherwise.
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And the range can be found either from the graph or the completed square form.
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When π is positive, the range will be π of π₯ is greater than or equal to π.
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And when π is negative, the range will be π of π₯ is less than or equal to π.
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In this video then, weβve seen how we can use either the graph or the equation of a quadratic function to determine these key features.