Question

question 2
given the function $f(x) = -2x^2$
find the difference quotient $\frac{f(x+h)-f(x)}{h}$
find $f(x)$ by determining $lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$.
find $f(-2)$
find $f(0)$
find $f(1)$
graph the function $f(x) = -2x^2$ and draw the tangent lines to the graph at points whose x-coordinates are -2, 0, and 1. (the slopes of these lines should match the derivative values you calculated above.)

Explanation:

Step1: Compute $f(x+h)$

$f(x+h) = -2(x+h)^2 = -2(x^2+2xh+h^2) = -2x^2-4xh-2h^2$

Step2: Calculate difference quotient

Substitute $f(x+h)$ and $f(x)$:

$$\begin{align*} \frac{f(x+h)-f(x)}{h}&=\frac{(-2x^2-4xh-2h^2)-(-2x^2)}{h}\\ &=\frac{-4xh-2h^2}{h}\\ &=-4x-2h \end{align*}$$

Step3: Find $f'(x)$ via limit

Take $\lim_{h\to0}$ of the quotient:

$$ f'(x)=\lim_{h\to0}(-4x-2h)=-4x $$

Step4: Compute $f'(-2)$

Substitute $x=-2$ into $f'(x)$:
$f'(-2)=-4(-2)=8$

Step5: Compute $f'(0)$

Substitute $x=0$ into $f'(x)$:
$f'(0)=-4(0)=0$

Step6: Compute $f'(1)$

Substitute $x=1$ into $f'(x)$:
$f'(1)=-4(1)=-4$

Answer:

Difference quotient: $\boldsymbol{-4x-2h}$
$f'(x)$: $\boldsymbol{-4x}$
$f'(-2)$: $\boldsymbol{8}$
$f'(0)$: $\boldsymbol{0}$
$f'(1)$: $\boldsymbol{-4}$

For the graph:

1. Plot $f(x)=-2x^2$, a downward-opening parabola with vertex at $(0,0)$.
2. At $x=-2$ (point $(-2, -8)$), draw a tangent line with slope $8$.
3. At $x=0$ (point $(0,0)$), draw a horizontal tangent line (slope $0$).
4. At $x=1$ (point $(1, -2)$), draw a tangent line with slope $-4$.