What is the vertices 4(x+3/2)^2+15 ?

The vertices 4(x+3/2)^2+15 is Minimum (-3/2 ,15)

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vertices 4(x+3/2)^2+15

vertices

4 (x + \frac{3}{2})^{2} + 15

Minimum (- \frac{3}{2}, 15)

Solution steps

y = 4 (x + \frac{3}{2})^{2} + 15

Rewrite

y = 4 (x + \frac{3}{2})^{2} + 15

in the form

y = a x^{2} + b x + c

y = 4 x^{2} + 12 x + 24

The parabola parameters are:

a = 4, b = 12, c = 24

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{12}{2 \cdot 4}

Simplify

- \frac{12}{2 \cdot 4} : - \frac{3}{2}

x_{v} = - \frac{3}{2}

Plug in

x_{v} = - \frac{3}{2}

to find the

y_{v}

value

y_{v} = 15

Therefore the parabola vertex is

(- \frac{3}{2}, 15)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (- \frac{3}{2}, 15)

f (x) = 2 x^{2} - 8 x

x^{2} + 4 x + 4 y - 4 = 0

y^{2} = - 24 x

f (x) = 2 x^{2} - 12 x + 23

x^{2} + 6 x + 7

What is the vertices 4(x+3/2)^2+15 ?
The vertices 4(x+3/2)^2+15 is Minimum (-3/2 ,15)