How do you write y=-x^2+20x-80y=−x2+20x−80 in vertex form?

2 Answers
Mia
Apr 26, 2017

y=-(x-10)^2+20" "y=−(x−10)2+20 is the vetex form.

Explanation:

The vertex form of a quadratic equation is :
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color(blue)(y=a(x-h)^2+k)" "y=a(x−h)2+k where (h,k) " "(h,k) is the vertex
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The quadratic form is performed by factorization.
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y=-x^2+20x-80y=−x2+20x−80
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rArry=-(x^2-20x+80)⇒y=−(x2−20x+80)
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rArry=-((x)^2-2(x)(10)+80)⇒y=−((x)2−2(x)(10)+80)
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To complete the square of the above equation we recognize that the second term should be color(red)(10^2=100102=100 so we should add 2020
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rArry=-((x)^2-2(x)(10)+80color(red)(+20-20))⇒y=−((x)2−2(x)(10)+80+20−20)
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rArry=-((x)^2-2(x)(10)+100-20)⇒y=−((x)2−2(x)(10)+100−20)
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rArry=-((x)^2-2(x)(10)+100)+20⇒y=−((x)2−2(x)(10)+100)+20
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rArry=-((x)^2-2(x)(10)+10^2)+20⇒y=−((x)2−2(x)(10)+102)+20
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rArry=-(x-10)^2+20⇒y=−(x−10)2+20
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Therefore,y=-(x-10)^2+20" "y=−(x−10)2+20 is the vertex form of the parabola where " "(10,20) (10,20) is its vertex.

EZ as pi
Apr 26, 2017

y = -(x-10)^2 +20y=−(x−10)2+20

Explanation:

To write a quadratic trinomial in vertex from you need to change:

y = ax^2 +bx +c " "y=ax2+bx+c into the form " "y = p(x+q)^2 + t y=p(x+q)2+t

In y = p(x+q)^2 + ty=p(x+q)2+t the vertex is at (-q, t)(−q,t)

y = -x^2 +20x-80" "larry=−x2+20x−80 ← make x^2x2 positive

y = -[x^2 -20x+80]" "larry=−[x2−20x+80] ← complete the square

y = -[x^2 -20x color(blue)(+ 100-100)+80]" "larrcolor(blue)(+(b/2)^2 -(b/2)^2)y=−[x2−20x+100−100+80] ←+(b2)2−(b2)2

y =-[(x-10)^2 -20]y=−[(x−10)2−20]

y = -(x-10)^2 +20y=−(x−10)2+20

This is now vertex form, giving the vertex as (10,20)(10,20)

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