WEBVTT 00:00:02.990 --> 00:00:09.820 Find the maximum value of the function 𝑦 is equal to negative two π‘₯ squared plus 12π‘₯ minus 25. 00:00:10.180 --> 00:00:13.450 In this question, we’re given a function 𝑦 which is a quadratic function. 00:00:13.490 --> 00:00:16.130 And we’re asked to find the maximum value of this function. 00:00:16.320 --> 00:00:19.740 We might be tempted to try applying the extreme value theorem to this question. 00:00:19.850 --> 00:00:26.110 However, the extreme value theorem is not applicable because we’re not looking for a maximum value over a closed interval. 00:00:26.210 --> 00:00:30.780 Instead, we need to find the maximum value of this function over the entire set of real numbers. 00:00:30.810 --> 00:00:33.810 To do this, let’s start by considering a sketch of our function. 00:00:33.970 --> 00:00:40.750 Since our function is a quadratic with a negative leading coefficient, we know it will have a parabola shape facing downwards. 00:00:40.840 --> 00:00:45.380 And the maximum value of this function will be at the vertex or turning point of this function. 00:00:45.500 --> 00:00:49.850 And in fact, we can prove this result and find this value in two different ways. 00:00:49.990 --> 00:00:53.850 We can either complete the square or we can use calculus. 00:00:53.990 --> 00:00:55.620 Let’s start by using calculus. 00:00:55.660 --> 00:01:00.430 First, we recall the local extrema of a function must occur at their critical points. 00:01:00.640 --> 00:01:07.680 And we also recall the critical points of a function are where the derivative is equal to zero or where the derivative does not exist. 00:01:07.820 --> 00:01:14.810 And in our case, because 𝑦 is a quadratic function, we know it’s continuous for all real values of π‘₯ and differentiable for all real values of π‘₯. 00:01:14.940 --> 00:01:20.540 So the only critical point of this function will be where its derivative is equal to zero, the maximum value of our function. 00:01:20.780 --> 00:01:25.370 So to find the maximum value of this function, we need to differentiate this with respect to π‘₯. 00:01:25.370 --> 00:01:28.770 We can do this term by term by using the power rule for differentiation. 00:01:28.960 --> 00:01:32.000 We multiply by the exponent of π‘₯ and reduce this exponent by one. 00:01:32.060 --> 00:01:34.790 Applying this to the first term, we get negative four π‘₯. 00:01:34.860 --> 00:01:40.640 And although we could apply this to evaluate the derivative of the next two terms, we notice these two terms are a linear function. 00:01:40.920 --> 00:01:48.960 And we recall the derivative of a linear function is the coefficient of the variable, in this case 12. d𝑦 by dπ‘₯ is negative four π‘₯ plus 12. 00:01:49.160 --> 00:01:51.930 Now, we’re ready to find the only critical point of our function. 00:01:51.960 --> 00:01:54.430 It will be when the derivative is equal to zero. 00:01:54.700 --> 00:01:58.320 So we need to solve the equation negative four π‘₯ plus 12 is equal to zero. 00:01:58.350 --> 00:02:03.700 We subtract 12 from both sides of the equation and divide through by negative four to get that π‘₯ is equal to three. 00:02:03.970 --> 00:02:06.720 Finally, we’re asked to find the maximum value of this function. 00:02:06.750 --> 00:02:10.310 So we need to substitute π‘₯ is equal to three into our function. 00:02:10.560 --> 00:02:16.270 This gives us that 𝑦 is equal to negative two times three squared plus 12 times three minus 25. 00:02:16.490 --> 00:02:20.270 And if we evaluate this expression, we see it’s equal to negative seven. 00:02:20.320 --> 00:02:26.690 The maximum value of the function 𝑦 is equal to negative two π‘₯ squared plus 12π‘₯ minus 25 is negative seven. 00:02:26.990 --> 00:02:29.900 However, this is not the only way we could have answered this question. 00:02:29.930 --> 00:02:32.200 We could have also used completing the square. 00:02:32.330 --> 00:02:37.680 We want to complete the square on the quadratic negative two π‘₯ squared plus 12π‘₯ minus 25. 00:02:37.890 --> 00:02:43.100 We’ll start by taking the constant factor of negative two outside of our first two terms. 00:02:43.340 --> 00:02:48.570 This gives us negative two multiplied by π‘₯ squared minus six π‘₯ minus 25. 00:02:48.910 --> 00:02:54.830 Next, we want to write π‘₯ squared minus six π‘₯ in the form π‘₯ plus π‘Ž all squared. 00:02:54.940 --> 00:02:58.440 And to do this, we recall we need to halve the coefficient of π‘₯. 00:02:58.600 --> 00:03:06.680 Therefore, if we distribute the exponent over π‘₯ minus three all squared, we have π‘₯ minus three all squared is π‘₯ squared minus six π‘₯ plus nine. 00:03:06.860 --> 00:03:12.240 We can then subtract nine from both sides of the equation to get an expression for π‘₯ squared minus six π‘₯. 00:03:12.430 --> 00:03:16.890 We have π‘₯ squared minus six π‘₯ is π‘₯ minus three all squared minus nine. 00:03:17.170 --> 00:03:19.650 We can then substitute this into our expression. 00:03:19.930 --> 00:03:26.400 This gives us negative two multiplied by π‘₯ minus three all squared minus nine minus 25. 00:03:26.660 --> 00:03:34.520 If we distribute the factor of negative two over our parentheses, we get negative two times π‘₯ minus three all squared plus 18 minus 25. 00:03:34.790 --> 00:03:36.080 And we can simplify this. 00:03:36.110 --> 00:03:38.900 18 minus 25 is equal to negative seven. 00:03:39.150 --> 00:03:46.220 And therefore, by completing the square, we’ve rewritten our function as negative two times π‘₯ minus three all squared minus seven. 00:03:46.420 --> 00:03:48.970 And we can use this to find the turning point of our parabola. 00:03:49.030 --> 00:03:51.240 But let’s start by recalling why this is true. 00:03:51.500 --> 00:03:54.470 We see that the expression π‘₯ minus three is squared. 00:03:54.500 --> 00:03:58.250 And the square of a number is always greater than or equal to zero. 00:03:58.650 --> 00:04:06.550 But then, we multiply this number by negative two, which means the entire first term of this expression is always less than or equal to zero. 00:04:06.880 --> 00:04:14.330 And this means to make this value as large as possible, we should try and make our first term equal to zero, since this is the biggest our first time can be. 00:04:14.630 --> 00:04:16.910 And this happens when π‘₯ is equal to three. 00:04:17.220 --> 00:04:21.290 Therefore, π‘₯ is equal to three is where the maximum value of this quadratic occurs. 00:04:21.320 --> 00:04:25.080 We can substitute it into our function and we will also get negative seven. 00:04:25.350 --> 00:04:33.560 And of course, since this is the value where our first term is equal to zero, the maximum value of our function is the second term, the constant value of negative seven. 00:04:33.830 --> 00:04:42.710 Therefore, we’ve seen two different ways of finding the maximum value of the quadratic function 𝑦 is equal to negative two π‘₯ squared plus 12π‘₯ minus 25. 00:04:42.770 --> 00:04:45.310 In both cases, this value was negative seven.