# How do you find the vertex and the intercepts for f(x)=x^2+x-2?

Mar 26, 2016

$V e r t e x \left(- \frac{1}{2} , - \frac{9}{4}\right)$

#### Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{1}{2}$
y-coordinate of vertex:
$y \left(- \frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} - 2 = - \frac{1}{4} - 2 = - \frac{9}{4}$
$V e r t e x \left(- \frac{1}{2} , - \frac{9}{4}\right)$.
To find y-intercept, make x = 0 --> y = -2
To find x-intercepts, make t = 0, and solve the quadratic equation:
${x}^{2} + x - 2 = 0.$
Since a + b + c = 0, use shortcut. The 2 x-intercepts (real roots) are:
1 and $\frac{c}{a} = - 2$.

Mar 26, 2016

$v \left(- \frac{1}{2} , - \frac{9}{4}\right)$

x int: $\left(- 2 , 0\right)$ and $\left(1 , 0\right)$

y int: $\left(0 , - 2\right)$

#### Explanation:

In a quadratic equation like $a {x}^{2} + b x + c$ , the formula $- \frac{b}{2 a}$ gives us the $x$ coordinate of the vertex. Then we put the $x$ value we found in the equation to get the $y$ value.

So, $x = - \frac{1}{2}$ $\to f \left(x\right) = y = \frac{1}{4} - \frac{1}{2} - 2 \to y = - \frac{9}{4} \to v \left(- \frac{1}{2} , - \frac{9}{4}\right)$

To find the $x$ intercepts, we should value $y$ as $0$ in the equation;

$0 = {x}^{2} + x - 2$
$0 = \left(x + 2\right) \left(x - 1\right)$
$x = - 2 , x = 1$

To find the $y$ intercept, we should value $x$ as $0$ in the equation;

$y = - 2$