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Popular >

range of 2x^2+15x+7

Solution

range

2 x^{2} + 15 x + 7

Solution

f (x) \geq - \frac{169}{8}

+1

Interval Notation

[- \frac{169}{8}, \infty)

Solution steps

Vertex of

2 x^{2} + 15 x + 7 :

Minimum

(- \frac{15}{4}, - \frac{169}{8})

For a parabola

a x^{2} + b x + c

with Vertex

(x_{v}, y_{v})

If

a < 0

the range is

f (x) \leq y_{v}

If

a > 0

the range is

f (x) \geq y_{v}

a = 2,

Vertex

(x_{v}, y_{v}) = (- \frac{15}{4}, - \frac{169}{8})

f (x) \geq - \frac{169}{8}

Popular Examples

inverse of f(x)=-8x+9 inverse

f (x) = - 8 x + 9

domain of 3x^2-8 domain

3 x^{2} - 8

domain of f(x)= 7/x+9/(x+9)domain

f (x) = \frac{7}{x} + \frac{9}{x + 9}

inverse of 1+sqrt(5+6x)inverse

1 + 5 + 6 x

range of f(x)=|x^2-4|+3 range

f (x) = x^{2} - 4 + 3

Frequently Asked Questions (FAQ)

What is the range of 2x^2+15x+7 ?
The range of 2x^2+15x+7 is f(x)>=-169/8