# How do you write  y = 2x^2 + 3x − 5 in vertex form and identify the vertex, y intercept and x intercept?

May 19, 2018

$V e r t e x = \left(- h , k\right) = \left(- \frac{3}{4} , - \frac{49}{8}\right)$

$\text{y-int} = - 5$

$\text{x-intercepts" = 1 " and } - \frac{5}{2}$

#### Explanation:

You must complete the square to solve this:

$y = 2 {x}^{2} + 3 x - 5$

$y + 5 = 2 {x}^{2} + 3 x$

vertex form is:

$y = a {\left(x - h\right)}^{2} + k$ where $\left(- h , k\right)$ is the vertex

to complete the square we need a leading coefficient of 1 so factor out the 2

$y + 5 + c = 2 \left({x}^{2} + \frac{3}{2} x + c\right)$

now $c = {\left(\frac{1}{2} \cdot \frac{3}{2}\right)}^{2} = {\left(\frac{3}{4}\right)}^{2} = \frac{9}{16}$

remember the c value must be multiplied by the 2 we factored out before we add it to the left side!

$y + 5 + \left(2 \cdot \frac{9}{16}\right) = 2 \left({x}^{2} + \frac{3}{2} x + \frac{9}{16}\right)$

$y + \frac{49}{8} = 2 {\left(x + \frac{3}{4}\right)}^{2}$

finally isolate the y and it is in vertex form:

$y = 2 {\left(x + \frac{3}{4}\right)}^{2} - \frac{49}{8}$

$V e r t e x = \left(- h , k\right) = \left(- \frac{3}{4} , - \frac{49}{8}\right)$

Y-int is $c$ for standard form:

$a {x}^{2} + b x + c$

$2 {x}^{2} + 3 x - 5$

$\text{y-int} = - 5$

you must factor it to find the roots (x-intercepts)

$y = 2 {x}^{2} + 3 x - 5$

$\left(x - 1\right) \left(2 x + 5\right)$

$x = 1 , - \frac{5}{2}$

So the x-intercepts are $1 , - \frac{5}{2}$