Question

graph the function.
$f(x) = -3x^2 + 1$
plot five points on the graph of the function: one point with $x = 0$, two points with negative $x$-values, and two points with positive $x$-values. then click on the graph-a-function button.

Explanation:

Step1: Find the point when \( x = 0 \)

Substitute \( x = 0 \) into \( f(x)=-3x^{2}+1 \).
\( f(0)=-3(0)^{2}+1 = 1 \). So the point is \( (0, 1) \).

Step2: Find points with negative \( x \)-values (e.g., \( x = -1, -2 \))

• For \( x = -1 \):

\( f(-1)=-3(-1)^{2}+1=-3 + 1=-2 \). Point: \( (-1, -2) \).

• For \( x = -2 \):

\( f(-2)=-3(-2)^{2}+1=-12 + 1=-11 \). Point: \( (-2, -11) \).

Step3: Find points with positive \( x \)-values (e.g., \( x = 1, 2 \))

• For \( x = 1 \):

\( f(1)=-3(1)^{2}+1=-3 + 1=-2 \). Point: \( (1, -2) \).

• For \( x = 2 \):

\( f(2)=-3(2)^{2}+1=-12 + 1=-11 \). Point: \( (2, -11) \).

Answer:

The five points are \( (0, 1) \), \( (-1, -2) \), \( (-2, -11) \), \( (1, -2) \), \( (2, -11) \). (Plot these points on the graph and then draw the parabola opening downward with vertex at \( (0, 1) \).)