How do you graph the parabola $y = 4 x^{2} + 8 x - 22$ $y = 4x^2 + 8x -22$ using vertex, intercepts and additional points?

Question

1783 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-19T14:55:22

UNDERSTAND THE EXPLANATION

$y = 4 x^{2} + 8 x - 22$ $y=4x^2+8x-22$

Taking out 4 as a common factor as 'a' is always a factor to be taken out

$y = 4 (x^{2} + 2 x - \frac{11}{2})$ $y=4(x^2+2x-11/2)$

$y = 4 (x^{2} + 2 x + 1 - \frac{11}{2} - 1)$ $y=4(x^2+2x+1-11/2-1)$

convert the equation in the form $y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

$y = 4 {(x + 1)}^{2} - \frac{13}{2}$ $y=4(x+1)^2-13/2$

I hope now can get vertex, x, and y-intercept AND axis of symmetry
if not then mention me in the comment for further help

How do you graph the parabola y=4x2+8x−22y = 4x^2 + 8x -22 using vertex, intercepts and additional points?