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Find the equations to the tangent lines of the curve π¦ equals π₯ plus eight multiplied by π₯ plus 10 at the points where this curve intersects the π₯-axis.
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Letβs begin, first of all, by determining the coordinates of these points.
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Weβre looking for the points where the curve intersects the π₯-axis.
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And we know that everywhere on the π₯-axis, π¦ is equal to zero.
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So by setting our equation for π¦ equal to zero, we have an equation which we can solve to find the π₯-coordinates of the points where the curve intersects the π₯-axis.
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This is a quadratic equation already in its factored form.
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So we can take each factor in turn, set it equal to zero, and then solve the resulting linear equation, giving π₯ plus eight equals zero, which leads to π₯ equals negative eight and π₯ plus 10 equals zero, which leads to π₯ equals negative 10.
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The two points at which this curve intersects the π₯-axis then are the points negative 10, zero and negative eight, zero.
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We can sketch this curve, if we want.
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Itβs a quadratic curve with a positive leading coefficient intersecting the π₯-axis at negative 10 and negative eight.
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So it looks a little something like this.
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Weβre then looking to find the equations of the lines, which are tangent to this curve at the points where it intersects the π₯-axis.
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So those are the two lines drawn in green.
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We know that the general equation of a straight line in its point-slope form is π¦ minus π¦ one equals π π₯ minus π₯ one, where π₯ one, π¦ one gives the coordinates of one point on the line and π gives its slope.
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We know the coordinates of one point on each of our tangent lines.
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So in order to apply this formula, we need to determine their slopes.
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We can do this by recalling that the slope of the line tangent to a curve at any given point is the same as the slope of the curve itself at that point, which we can find by evaluating its first derivative.
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Before we get into differentiating, letβs manipulate our expression for this curve slightly by distributing the parentheses and then simplifying to give π¦ is equal to π₯ squared plus 18π₯ plus 80, a polynomial.
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We recall then that the power rule of differentiation, which tells us that the derivative with respect to π₯ of a general power term ππ₯ to the πth power for real values of π and π is equal to ππ multiplied by π₯ to the power of π minus one.
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We multiply by the exponent and then reduce the exponent by one.
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We can now apply this power rule of differentiation to find the derivative of our function π¦.
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The derivative of π₯ squared with respect to π₯ is two π₯.
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The derivative of positive 18π₯ is positive 18, which we can see if we think of 18π₯ as 18π₯ to the power of one.
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And finally, the derivative of a constant term positive 80 is just zero, which we can see if we think of 80 as 80π₯ to the power of zero.
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We find then that the general slope function of our curve dπ¦ by dπ₯ is equal to two π₯ plus 18.
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We then need to evaluate the slope at each of the points weβre interested in.
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First of all, when π₯ is equal to negative 10, the slope will be equal to two multiplied by negative 10 plus 18.
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Thatβs negative 20 plus 18 which is equal to negative two.
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Secondly, when π₯ is equal to negative eight, the slope will be equal to two multiplied by negative eight plus 18.
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Thatβs negative 16 plus 18 which is equal to two.
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Notice that this is consistent with what weβve seen on our sketch.
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The tangent line at the point where π₯ equals negative 10 has a negative slope.
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Whereas the tangent to the point when π₯ equals negative eight has positive slope.
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Next, we can use the general equation of our straight line.
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At the point with coordinates negative 10, zero where the slope is equal to negative two, we have the equation π¦ minus zero equals negative two multiplied by π₯ minus negative 10.
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That simplifies to π¦ equals negative two multiplied by π₯ plus 10.
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And distributing the parentheses, we have π¦ equals negative two π₯ minus 20.
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We can then collect all the terms on the left-hand side of the equation to give the equation of this tangent as π¦ plus two π₯ plus 20 is equal to zero.
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For our other tangent at the point negative eight, zero where the slope is equal to two, we have the equation π¦ minus zero equals two multiplied by π₯ minus negative eight.
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That simplifies to π¦ equals two multiplied by π₯ plus eight.
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And then distributing the parentheses gives π¦ equals two π₯ plus 16.
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Again, collecting all terms on the left-hand side of the equation gives the equation of this tangent line in the form π¦ minus two π₯ minus 16 is equal to zero.
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So weβve completed the problem and found the equations of both lines, which are tangent to this curve at the points where it intersects the π₯-axis.
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They are π¦ plus two π₯ plus 20 equals zero and π¦ minus two π₯ minus 16 equals zero.