How do you find the vertex and axis of symmetry of $y = \frac{5}{4} {(x + 2)}^{2} - 1$ $y=5/4(x+2)^2 -1$ ?

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Answer 1 · 2017-06-28T14:02:01

Vertex is $(- 2, - 1)$ $(-2,-1)$ and axis of symmetry is $x + 2 = 0$ $x+2=0$ .

This is the equation of a horizontal parabola in vertex form i.e.

$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

where vertex is $(h, k)$ $(h,k)$ and axis of symmetry is $x = h$ $x=h$ .

Hence vertex and axis of symmetry of $y = \frac{1}{4} {(x + 2)}^{2} - 1$ $y=1/4(x+2)^2-1$

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

are $(- 2, - 1)$ $(-2,-1)$ and $x = - 2$ $x=-2$ or $x + 2 = 0$ $x+2=0$ respectively

graph{((x+2)^2-4-4y)(x+2)((x+2)^2+(y+1)^2-0.02)=0 [-13, 7, -4.28, 5.72]}

How do you find the vertex and axis of symmetry of y=54(x+2)2−1y=5/4(x+2)^2 -1?