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Popular >

p(x)=0.02x^2+0.6x+20

Solution

p (x) = 0.02 x^{2} + 0.6 x + 20

Solution

Domain : - \infty < x < \infty

Range : f (x) \geq 15.5

Y Intercepts : (0, 20)

Vertex : Minimum (- 15, 15.5)

Inverse : 5 (2 x - 31 - 3), - 5 (2 x - 31 + 3)

+1

Interval Notation

Domain : (- \infty, \infty)

Range : [15.5, \infty)

Solution steps

Domain of

0.02 x^{2} + 0.6 x + 20 : - \infty < x < \infty

Range of

0.02 x^{2} + 0.6 x + 20 : f (x) \geq 15.5

Axis interception points of

0.02 x^{2} + 0.6 x + 20 :

Y Intercepts

: (0, 20)

Vertex of

0.02 x^{2} + 0.6 x + 20 :

Minimum

(- 15, 15.5)

Inverse of

0.02 x^{2} + 0.6 x + 20 : 5 (2 x - 31 - 3), - 5 (2 x - 31 + 3)

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