Natasha Henry

2022-01-23

How do you find the vertex of $y=2{x}^{2}+10x+8$?

Kyler Jacobson

Expert

Step 1
Given the equation of parabola in standard form

then the x-coordinate of the vertex is
${x}_{vertex}=-\frac{b}{2a}$
$y=2{x}^{2}+10x+8$ is in standard form
with $a=2,b=10$ and $c=8$
$⇒{x}_{vertex}=-\frac{10}{4}=-\frac{5}{2}$
substitute this value the equation for y-coordinate
${y}_{vertex}=2{\left(-\frac{5}{2}\right)}^{2}+10\left(-\frac{5}{2}\right)+8=-\frac{9}{2}$

Graph:

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