Question

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.

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=- 2x^{2}$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=-4x$.

Step2: Find the slope of the tangent line at $x=-2$

Substitute $x = - 2$ into $f'(x)$: $f'(-2)=-4\times(-2)=8$.
When $x=-2$, $f(-2)=-2\times(-2)^{2}=-8$. The equation of the tangent line using the point - slope form $y - y_1=m(x - x_1)$ (where $(x_1,y_1)=(-2,-8)$ and $m = 8$) is $y+8=8(x + 2)$, or $y=8x + 8$.

Step3: Find the slope of the tangent line at $x = 0$

Substitute $x = 0$ into $f'(x)$: $f'(0)=-4\times0=0$.
When $x = 0$, $f(0)=-2\times0^{2}=0$. The equation of the tangent line is $y-0=0(x - 0)$, or $y = 0$.

Step4: Find the slope of the tangent line at $x = 1$

Substitute $x = 1$ into $f'(x)$: $f'(1)=-4\times1=-4$.
When $x = 1$, $f(1)=-2\times1^{2}=-2$. The equation of the tangent line using the point - slope form (where $(x_1,y_1)=(1,-2)$ and $m=-4$) is $y + 2=-4(x - 1)$, or $y=-4x+2$.

Step5: Graph the function and tangent lines

The function $y=-2x^{2}$ is a parabola opening downwards with vertex at the origin $(0,0)$. Plot the points and draw the lines $y = 8x + 8$, $y = 0$, and $y=-4x + 2$ on the same graph as the parabola.

Answer:

Graph the parabola $y=-2x^{2}$ and the lines $y = 8x + 8$, $y = 0$, $y=-4x+2$ on the given grid.