How do you find the vertex of a parabola in standard form?

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How do you find the vertex of a parabola in standard form?

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Answer 1 · 2018-08-11T00:32:34

Refer to the explanation.

Explanation:

The standard form of a parabola is $y = a x^{2} + + b x + c$ $y=ax^2++bx+c$ , where $a \neq 0$ $a!=0$ .

The vertex is the minimum or maximum point of a parabola. If $a > 0$ $a>0$ , the vertex is the minimum point and the parabola opens upward. If $a < 0$ $a<0$ , the vertex is the maximum point and the parabola opens downward.

To find the vertex, you need to find the x- and y-coordinates.

The formula for the axis of symmetry and the x-coordinate of the vertex is:

$x = \frac{- b}{2 a}$ $x=(-b)/(2a)$

To find the y-coordinate of the vertex, substitute the value for $x$ $x$ into the equation and solve for $y$ $y$ .

$y = a {(\frac{- b}{2 a})}^{2} + b (\frac{- b}{2 a}) + c$ $y=a((-b)/(2a))^2+b((-b)/(2a))+c$

Example:

Find the vertex of $y = x^{2} + 4 x - 9$ $y=x^2+4x-9$ , where: $a = 1$ $a=1$ , $b = 4$ $b=4$ , and $c = - 9$ $c=-9$ .

Step 1. Find the x-coordinate of the vertex

$x = \frac{- 4}{2 \cdot 1}$ $x=(-4)/(2*1)$

$x = - \frac{4}{2}$ $x=-4/2$

$x = - 2$ $x=-2$ $\leftarrow$ $larr$ x-coordinate of the vertex

Step 2. Find the y-coordinate of the vertex.

Substitute $- 2$ $-2$ for $x$ $x$ and solve for $y$ $y$ .

$y = {(- 2)}^{2} + 4 (- 2) - 9$ $y=(-2)^2+4(-2)-9$

$y = 4 - 8 - 9$ $y=4-8-9$

$y = - 13$ $y=-13$ $\leftarrow$ $larr$ y-coordinate of the vertex

The vertex is $(- 2, - 13)$ $(-2,-13)$ .

graph{y=x^2+4x-9 [-9.71, 10.29, -13.68, -3.68]}

Answer 2 · 2018-08-11T01:45:14

See the wholesome answer, in the explanation.

Explanation:

When the axis and the perpendicular tangent at the vertex

$V (α, β)$ $V ( alpha, beta )$

are parallel to the coordinate axes, the standard form is

${(y - β)}^{2} = \pm 4 a (x - α)$ $( y - beta )^2 = +- 4a ( x - alpha)$ or

${(x - α)}^{2} = \pm 4 a (y - β)$ $( x - alpha )^2 = +- 4a ( y - beta )$

Graph for

${(y + 3)}^{2} = - 4 (x - 1)$ $(y+3)^2 = - 4 ( x - 1 )$ , and vice versa, ${(x - 1)}^{2} = - 4 (y + 3)$ $( x - 1 )^2 = - 4 ( y + 3)$ :
graph{((y+3)^2 + 4 ( x - 1 ))(( x - 1 )^2 + 4 ( y + 3 ))((x-1)^2+(y+3)^2-0.01)=0[-5 7 -6 0]}

If ab = h^2, the general 2nd-degree equation

$a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a parabola,

and my standard form here is

${((y - β) - m (x - α))}^{2} = \pm 4 a (m (y - β) + (x - α))$ $((y-beta) - m( x - alpha ) )^2 = +-4a(m ( y-beta ) +( x - alpha ) )$ or

$\pm 4 a ((y - β) - m (x - α)) = {(m (y - β) + (x - α))}^{2}$ $+- 4a ((y-beta) - m( x - alpha ) ) = ( m ( y-beta ) +( x - alpha ) )^2$

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

Vertex : $V (α, β) = (- 2, 2)$ $V ( alpha, beta ) = ( - 2, 2 )$

Axis: $(y - 2) - 3 (x + 2) = 0$ $( y - 2 ) - 3 ( x + 2 ) = 0$

Tangent at V: $3 (y - 2) + (x + 2) = 0$ $3 ( y -2 ) + ( x + 2 ) = 0$

Graph, with the axis and the tangent at $V (- 2, 2)$ $V ( - 2, 2 )$ :
graph{(( ( y - 2 ) - 3 ( x + 2 ))^2 + 2 ( 3 ( y -2 ) + ( x + 2 )))(( y - 2 ) - 3 ( x + 2 ))( 3 ( y -2 ) + ( x + 2 )) = 0}

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How do you find the vertex of a parabola in standard form?

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