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

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Answer 1 · 2018-06-11T13:25:03

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

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

(0,7) is the intercept.

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

$- \frac{1}{2}$ $-1/2$

Put this into the equation

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

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

when x=0, y=7

when y=0, x= $\frac{- 1 \pm \sqrt{1 - 28}}{2}$ $[-1\pmsqrt(1-28)]/2$

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

How do you find the vertex, axis of symmetry and intercepts of f(x)=x^2+x+7f(x)=x2+x+7f(x)=x^2+x+7?