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Arlecino [84]
3 years ago
11

What is the vertex of the quadratic function f(x)=x2+6x+16? Write your answer

Mathematics
1 answer:
Olenka [21]3 years ago
5 0

Answer:

The vertex of the quadratic function is:

(x_{v}, y_{v})=\left(-3,\:7\right)

Step-by-step explanation:

Given the function

f\left(x\right)=x^2+6x+16

As the vertex of the form y=ax^2+bx+c is defined as:

x_v=-\frac{b}{2a}

As the quadratic function of parabola params are

a=1,\:b=6,\:c=16

so

x_v=-\frac{b}{2a}

x_v=-\frac{6}{2\cdot \:1}

x_v=-3

Putting x_v=-3 to determine y_v

y_v=\left(-3\right)^2+6\left(-3\right)+16

y_v=3^2-18+16

y_v=9-2

y_v=7

Therefore, the vertex of the quadratic function is:

(x_{v}, y_{v})=\left(-3,\:7\right)

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(y - 257/74)^2 = 1817/5476

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y - 257/74 = sqrt(1817)/74 or y - 257/74 = -sqrt(1817)/74

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Subtract 4 from both sides:

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Multiply both sides by -1:

x = 39/74 - sqrt(1817)/74 or y - 257/74 = -sqrt(1817)/74

Add 257/74 to both sides:

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Subtract 4 from both sides:

x = 39/74 - sqrt(1817)/74 or -x = -39/74 - sqrt(1817)/74

Multiply both sides by -1:

Answer:  x = 39/74 - sqrt(1817)/74 or x = 39/74 + sqrt(1817)/74

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