given the function $f(x)=-2x^{2}$\nfind the difference quotient $\frac{f(x + h)-f(x)}{h}$\nfind $f(x)$ by…

given the function $f(x)=-2x^{2}$\nfind the difference quotient $\frac{f(x + h)-f(x)}{h}$\nfind $f(x)$ by determining $lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$\nuse the previous answer to find the values indicated below. the values should match the slopes of the tangent lines you will draw on the graph at the bottom.\nfind $f(-2)$\nfind $f(0)$\nfind $f(1)$\ngraph the function $f(x)=-2x^{2}$ and draw the tangent lines to the graph at points whose x - coordinates are - 2, 0, and 1.
Answer
Explanation:
Step1: Find the difference quotient
First, find (f(x + h)): [ \begin{align*} f(x+h)&=-2(x + h)^{2}\ &=-2(x^{2}+2xh+h^{2})\ &=-2x^{2}-4xh - 2h^{2} \end{align*} ] Then, calculate (\frac{f(x + h)-f(x)}{h}): [ \begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{-2x^{2}-4xh - 2h^{2}-(-2x^{2})}{h}\ &=\frac{-2x^{2}-4xh - 2h^{2}+2x^{2}}{h}\ &=\frac{-4xh - 2h^{2}}{h}\ &=-4x-2h \end{align*} ]
Step2: Find the derivative (f^{\prime}(x))
[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}\ &=\lim_{h\rightarrow0}(-4x - 2h)\ &=-4x \end{align*} ]
Step3: Find (f^{\prime}(-2))
Substitute (x=-2) into (f^{\prime}(x)): [f^{\prime}(-2)=-4\times(-2)=8]
Step4: Find (f^{\prime}(0))
Substitute (x = 0) into (f^{\prime}(x)): [f^{\prime}(0)=-4\times0 = 0]
Step5: Find (f^{\prime}(1))
Substitute (x = 1) into (f^{\prime}(x)): [f^{\prime}(1)=-4\times1=-4]
Answer:
Difference quotient: (-4x - 2h) (f^{\prime}(x)): (-4x) (f^{\prime}(-2)): (8) (f^{\prime}(0)): (0) (f^{\prime}(1)): (-4)