Exercise 36 Page 210

For a quadratic function $f (x) = a x^{2} + b x + c,$ the $y -$ coordinate of the vertex is the minimum value of the function when $a > 0 .$

Let's simplify the equation of the given function to identify

a,

b,

and

c .

f (x) = 3 (x - 10) (x + 4)

▼

Multiply parentheses

Distr

Distribute $3$

f (x) = (3 x - 30) (x + 4)

Distr

Distribute $(x + 4)$

f (x) = 3 x (x + 4) - 30 (x + 4)

Distr

Distribute $3 x$

f (x) = 3 x^{2} + 12 x - 30 (x + 4)

Distr

Distribute $- 30$

f (x) = 3 x^{2} + 12 x - 30 x - 120

SubTerms

Subtract terms

f (x) = 3 x^{2} - 18 x - 120

DiffToSum

$a - b = a + (- b)$

f (x) = 3 x^{2} + (- 18) x + (- 120)

We can see above that $a = 3,$ $b = - 18,$ and $c = - 120 .$ We will now use these values to find the desired information.

Minimum Value

Since

a = 3

is greater than

0,

the parabola will open upwards. This means it will have a minimum value, which is given by

f (- \frac{b}{2 a}) .

Before we find the value of the function at this point, we need to substitute

a = 3

and

b = - 18

in

- \frac{b}{2 a} .

- \frac{b}{2 a}

▼

Substitute values and evaluate

SubstituteII

$a = 3$ , $b = - 18$

- \frac{- 1 8}{2 ( 3 )}

Multiply

- \frac{- 1 8}{6}

CalcQuot

Calculate quotient

- (- 3)

NegNeg

$- (- a) = a$

3

Now we have to calculate

f (3) .

To do so, we will substitute

3

for

x

in the given function.

f (x) = 3 (x - 10) (x + 4)

Substitute

$x = 3$

f (3) = 3 (3 - 10) (3 + 4)

AddSubTerms

Add and subtract terms

f (3) = 3 (- 7) (7)

Multiply

f (3) = - 147

This tells us that the minimum value of the function is

- 147 .

Domain and Range

Unless there are any specified restrictions on the

x -

values, the domain of a quadratic function is all real numbers. Therefore, the domain of this function is all real numbers. Furthermore, since

a = 3

is greater than

0,

the range is all values greater than or equal to the minimum value,

- 147 .

Domain : Range : All real numbers y \geq - 147

Decreasing and Increasing Intervals

Since

a = 3

is greater than

0,

the function decreases to the left of the minimum value and increases to the right of the minimum value, which we know occurs at

x = 3 .

Decreasing Interval : Increasing Interval : To the left of 3 To the right of 3