Question

graphing quadratics from standard form handout1)quadratic equation$y = 2x^2 + 8x + 5$identify a, b, and c$a=LXB0c=$find the vertex (x, y)$x = \frac{-b}{2a}$substitute x to find y...vertex: (_, _)graph

Explanation:

Step1: Identify a, b, c values

Standard quadratic form: $y=ax^2+bx+c$. For $y=2x^2+8x+5$,
$a=2$, $b=8$, $c=5$

Step2: Calculate x-coordinate of vertex

Use formula $x=\frac{-b}{2a}$
$x=\frac{-8}{2\times2}=\frac{-8}{4}=-2$

Step3: Calculate y-coordinate of vertex

Substitute $x=-2$ into $y=2x^2+8x+5$
$y=2(-2)^2+8(-2)+5=2(4)-16+5=8-16+5=-3$

Step4: Prepare graph points

Vertex: $(-2,-3)$. Additional points:
When $x=-1$, $y=2(-1)^2+8(-1)+5=2-8+5=-1$
When $x=0$, $y=2(0)^2+8(0)+5=5$
When $x=-3$, $y=2(-3)^2+8(-3)+5=18-24+5=-1$
When $x=-4$, $y=2(-4)^2+8(-4)+5=32-32+5=5$

Answer:

Identify a, b, and c:
$a=2$, $b=8$, $c=5$
Vertex: $(-2, -3)$
Graph: Plot the vertex $(-2,-3)$ and additional points $(-1,-1)$, $(0,5)$, $(-3,-1)$, $(-4,5)$, then draw a symmetric parabola opening upwards.