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graph the function $f(x) = 2x^2 + 8x + 14$. plot the vertex. then plot …

Question

graph the function $f(x) = 2x^2 + 8x + 14$. plot the vertex. then plot another point on the parabola. if you make a mistake, you can erase your parabola by selecting the second point and placing it on top of the first.

Explanation:

Step1: Find vertex x-coordinate

For $f(x)=ax^2+bx+c$, $x_v = \frac{-b}{2a}$
$a=2$, $b=8$, so $x_v = \frac{-8}{2\times2} = -2$

Step2: Find vertex y-coordinate

Substitute $x=-2$ into $f(x)$:
$f(-2)=2(-2)^2+8(-2)+14 = 2(4)-16+14 = 8-16+14 = 6$
Vertex is $(-2, 6)$

Step3: Find another point

Choose $x=0$:
$f(0)=2(0)^2+8(0)+14 = 14$
Point is $(0, 14)$

Answer:

  1. Plot the vertex at $(-2, 6)$
  2. Plot the second point at $(0, 14)$

(The parabola opens upward, symmetric about the line $x=-2$)