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KonstantinChe [14]
3 years ago
15

Write x^2-12x+8 in vertex form.

Mathematics
1 answer:
solong [7]3 years ago
8 0
The vertex form: y = a(x - h)² + k

y = ax² + bx + c then h = -b/2a and k = f(h).

y = x² - 12x + 8

a = 1; b = -12; c = 8

h = -(-12)/(2·1) = 12/2 = 6

k = f(6) = 6² - 12·6 + 8 = 36 - 72 + 8 = -28

Answer: y = (x - 6)² - 28.
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Determine whether f(x) = –5x^2 – 10x + 6 has a maximum or a minimum value. Find that value and explain how you know.
polet [3.4K]

Answer:

The vertex is the point (-1,11). is a maximum

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

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

where

(h,k) is the vertex

a is a coefficient

if a > 0 the parabola open upward and the vertex is a minimum

if a < 0 the parabola open downward and the vertex is a maximum

we have

f(x)=-5x^{2}-10x+6

Convert to vertex form

Complete the square

f(x)-6=-5x^{2}-10x

Factor the leading coefficient

f(x)-6=-5(x^{2}+2x)

f(x)-6-5=-5(x^{2}+2x+1)

f(x)-11=-5(x^{2}+2x+1)

Rewrite as perfect squares

f(x)-11=-5(x+1)^{2}

f(x)=-5(x+1)^{2}+11

The vertex is the point (-1,11)

The coefficient a=-5

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a < 0 the parabola open downward and the vertex is a maximum

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