# How do you find the vertex and the intercepts for  y=2x^2 + 8x + 5?

Oct 21, 2017

Vertex: $\left(- 3 , - 2\right)$

$x$-intercepts:

$x = - 3.2247448713916 , \setminus q \quad \setminus q \quad \setminus q \quad x = - 0.77525512860841$

$y$-intercept: $\left(0 , 5\right)$

#### Explanation:

Since the quadratic equation is given in standard form, we can find the vertex and roots using the vertex and quadratic formula.

Standard form is $a {x}^{2} + b x + c$. In this case:

$a = 2$

$b = 8$

$c = 5$

The vertex formula for the $x$-coordinate is $- \setminus \frac{b}{2 a}$

Plugging in yields:

$- \setminus \frac{8}{2 \setminus \cdot 2}$

$= - 2$

To find the $y$-coordinate, plug in the $x$-coordinate ($- 2$) into the original quadratic equation, in place of $x$:

$2 {x}^{2} + 8 x + 5$

$\setminus R i g h t a r r o w 2 {\left(- 2\right)}^{2} + 8 \left(- 2\right) + 5$

$\setminus R i g h t a r r o w 2 \left(4\right) - 16 + 5$

$\setminus R i g h t a r r o w 8 - 16 + 5$

$\setminus R i g h t a r r o w - 3$

Therefore, the vertex is $\left(- 2 , - 3\right)$

Now, to find the $x$-intercepts, plug the corresponding $a$, $b$, and $c$ values into the quadratic formula

$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \setminus \frac{- 8 \setminus \pm \setminus \sqrt{{8}^{2} - 4 \left(2\right) \left(5\right)}}{2 \left(2\right)}$

$x = \setminus \frac{- 8 \setminus \pm \setminus \sqrt{64 - 40}}{4}$

$x = \setminus \frac{- 8 \setminus \pm \setminus \sqrt{24}}{4}$

$x = \setminus \frac{- 8 \setminus \pm 2 \setminus \sqrt{6}}{4}$

Splitting $x$ up into the plus and minus values:

$x = \setminus \frac{- 8 + 2 \setminus \sqrt{6}}{4} , \setminus q \quad \setminus q \quad \setminus q \quad x = \setminus \frac{- 8 - 2 \setminus \sqrt{6}}{4}$

$x = - 3.2247448713916 , \setminus q \quad \setminus q \quad \setminus q \quad x = - 0.77525512860841$

Those are the $x$intercepts.

The $y$-intercept is simply $\left(0 , c\right)$, or $\left(0 , 5\right)$ in this case.