Question

consider the parabola y = 8x - x². (a) find the slope of the tangent line to the parabola at the point (1, 7). (b) find an equation of the tangent line in part (a). (c) graph the parabola and the tangent line.

Explanation:

Step1: Differentiate the function

Given $y = 8x - x^{2}$, using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $y'=\frac{d}{dx}(8x - x^{2})=8-2x$.

Step2: Find the slope at the given point

Substitute $x = 1$ into $y'$. When $x = 1$, $y'(1)=8-2\times1=6$. So the slope of the tangent line at the point $(1,7)$ is 6.

Step3: Find the equation of the tangent line

Use the point - slope form of a line $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,7)$ and $m = 6$.
$y-7=6(x - 1)$
Expand: $y-7=6x-6$
Rearrange to get $y=6x + 1$.

Answer:

(a) 6
(b) $y = 6x+1$
(c) To graph the parabola $y = 8x - x^{2}=-(x^{2}-8x)=-(x - 4)^{2}+16$, it is a parabola opening downwards with vertex at $(4,16)$. The tangent line $y = 6x + 1$ has a $y$ - intercept of 1 and a slope of 6. You can use a graphing utility (such as a graphing calculator or software like Desmos) to accurately graph both the parabola and the tangent line. The parabola and the tangent line should intersect at the point $(1,7)$.