What is the axis of symmetry and vertex for the graph $y = - x^{2} - 3 x + 2$ $y=-x^2-3x+2$ ?

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Answer 1 · 2017-03-19T11:14:30

The axis of symmetry is $x = - \frac{3}{2}$ $x=-3/2$
The vertex is $= (- \frac{3}{2}, \frac{17}{4})$ $=(-3/2,17/4)$

We use

$a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ $a^2-2ab+b^2=(a-b)^2$

We complete the square and factorise in order to find the vertex form.

$y = - x^{2} - 3 x + 2$ $y=-x^2-3x+2$

$y = - (x^{2} + 3 x) + 2$ $y=-(x^2+3x)+2$

$y = - (x^{2} + 3 x + \frac{9}{4}) + 2 + \frac{9}{4}$ $y=-(x^2+3x+9/4)+2+9/4$

$y = - {(x + \frac{3}{2})}^{2} + \frac{17}{4}$ $y=-(x+3/2)^2+17/4$

This is the vertex form of the equation.

The axis of symmetry is $x = - \frac{3}{2}$ $x=-3/2$

The vertex is $= (- \frac{3}{2}, \frac{17}{4})$ $=(-3/2,17/4)$

graph{(y+(x+3/2)^2-17/4)((x+3/2)^2+(y-17/4)^2-0.02)(y-1000(x+3/2))=0 [-11.25, 11.25, -5.625, 5.625]}

What is the axis of symmetry and vertex for the graph y=-x^2-3x+2y=−x2−3x+2y=-x^2-3x+2?