# How do you find the vertex, axis of symmetry and intercepts of f(x)=x^2+x+7?

Jun 11, 2018

vertex $\left(- \frac{1}{2} , 6 \frac{3}{4}\right)$

$x = - \frac{1}{2}$ is the axis of symmetry

(0,7) is the intercept.

#### Explanation:

$\frac{- b}{2 a}$ gives the x coordinate for the vertex

$- \frac{1}{2}$

Put this into the equation

$y = \frac{1}{4} - \frac{1}{2} + 7$

$y = 6 \frac{3}{4}$

when x=0, y=7

when y=0, x=$\frac{- 1 \setminus \pm \sqrt{1 - 28}}{2}$

$x = \frac{- 1 \setminus \pm \sqrt{-} 27}{2}$ cannot have a negative square root so no solutions