consider the parabola y = 8x - x². (a) find the slope of the tangent line to the parabola at the point (1…

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 $(1,7)$.