Question

(a) find the slope of the tangent line to the parabola y = x² + 7x at the point (-3, -12) by using the following parameters.
(i) the tangent line to the curve y = f(x) at the point p(a, f(a)) is the line through p with slope m = lim(x→a) (f(x)-f(a))/(x - a) provided that this limit exists.
m =
(ii) the expression of slope is m = lim(h→0) (f(a + h)-f(a))/h.
m =
(b) find an equation of the tangent line in part (a).
y =
(c) graph the parabola and the tangent line. as a check on your work, zoom in toward the point (-3, -12) until the parabola and the tangent line are indistinguishable.

Explanation:

Step1: Identify the function and the point

The function is $f(x)=x^{2}+7x$ and the point is $a = - 3$.

Step2: Use the first - slope formula $\lim_{x

ightarrow a}\frac{f(x)-f(a)}{x - a}$
First, find $f(-3)=(-3)^{2}+7\times(-3)=9 - 21=-12$. Then $f(x)-f(-3)=x^{2}+7x+12$. Factor $x^{2}+7x + 12=(x + 3)(x+4)$. So $\lim_{x
ightarrow - 3}\frac{f(x)-f(-3)}{x+3}=\lim_{x
ightarrow - 3}\frac{(x + 3)(x + 4)}{x+3}=\lim_{x
ightarrow - 3}(x + 4)=1$.

Step3: Use the second - slope formula $\lim_{h

ightarrow0}\frac{f(a + h)-f(a)}{h}$
$f(-3+h)=(-3 + h)^{2}+7(-3 + h)=9-6h+h^{2}-21 + 7h=h^{2}+h - 12$. And $f(-3)=-12$. Then $\lim_{h
ightarrow0}\frac{f(-3 + h)-f(-3)}{h}=\lim_{h
ightarrow0}\frac{h^{2}+h-12 + 12}{h}=\lim_{h
ightarrow0}\frac{h^{2}+h}{h}=\lim_{h
ightarrow0}(h + 1)=1$.

Step4: Find the equation of the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-3,-12)$ and $m = 1$. So $y+12=1\times(x + 3)$, which simplifies to $y=x - 9$.

Answer:

(a)(i) $1$
(a)(ii) $1$
(b) $y=x - 9$
(c) The correct graph is the one where the line $y=x - 9$ touches the parabola $y=x^{2}+7x$ at the point $(-3,-12)$.