Question

PLEASE HELP

1) What is the minimum value for g(x)=x^2−10x+16? Enter your answer in the box.

2) What are the x-intercepts of the quadratic function? f(x)=x2−3x−10
Enter your answers in the boxes. __and__

3) Let ​ f(x)=x2+17x+72 ​ What are the zeros of the function?
​Enter your answers in the boxes. ​___and___

4) Let ​ f(x)=x2+6x+11 ​. What is the minimum value of the function?
Enter your answer in the box. ​___and___

Answer 1

-31

Explanation:

Answer 2

Part 1) The minimum value is
`(5,-9)`

Part 2) The x-intercepts are -2 and 5

Part 3) The zeros of the function are -9 and -8

Part 4) The minimum value is
`(-3,2)`

Explanation:

Part 1) we have

`g(x)=x^(2)-10x+16`

we know that

The equation of a vertical parabola in vertex form is equal to

`y=a(x-h)^(2)+k`

where

(h,k) is the vertex

if a>0---> the the parabola open upward (vertex is a minimum)

If a<0---> the the parabola open downward (vertex is a maximum)

Convert to vertex form

`g(x)-16=x^(2)-10x`

`g(x)-16+25=x^(2)-10x+25`

`g(x)+9=x^(2)-10x+25`

`g(x)+9=(x-5)^(2)`

`g(x)=(x-5)^(2)-9` ------> vertex form

The vertex is the point
`(5,-9)`

the parabola open upward (vertex is a minimum)

Part 2) we have

`f(x)=x^(2)-3x-10`

we know that

The x-intercepts are the values of x when the value of the function is equal to zero

so

equate the equation to zero

`x^(2)-3x-10=0`

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`x^(2)-3x-10=0`

so

`a=1\\b=-3\\c=-10`

substitute in the formula

`x=\frac{-(-3)(+/-)\sqrt{-3^(2)-4(1)(-10)}} {2(1)}`

`x=\frac{3(+/-)√(49)} {2}`

`x=\frac{3(+/-)7} {2}`

`x=\frac{3(+)7} {2}=5`

`x=\frac{3(-)7} {2}=-2`

Part 3) we have

`f(x)=x^(2)+17x+72`

we know that

The zeros of the function are the values of x when the value of the function is equal to zero

so

equate the equation to zero

`x^(2)+17x+72=0`

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`x^(2)+17x+72=0`

so

`a=1\\b=17\\c=72`

substitute in the formula

`x=\frac{-17(+/-)\sqrt{17^(2)-4(1)(72)}} {2(1)}`

`x=\frac{-17(+/-)√(1)} {2}`

`x=\frac{-17(+/-)1} {2}`

`x=\frac{-17(+)1} {2}=-8`

`x=\frac{-17(-)1} {2}=-9`

Part 4) we have

`f(x)=x^(2)+6x+11`

we know that

The equation of a vertical parabola in vertex form is equal to

`y=a(x-h)^(2)+k`

where

(h,k) is the vertex

if a>0---> the the parabola open upward (vertex is a minimum)

If a<0---> the the parabola open downward (vertex is a maximum)

Convert to vertex form

`f(x)-11=x^(2)+6x`

`f(x)-11+9=x^(2)+6x+9`

`f(x)-2=x^(2)+6x+9`

`f(x)-2=(x+3)^(2)`

`f(x)=(x+3)^(2)+2`

The vertex is the point
`(-3,2)`

the parabola open upward (vertex is a minimum)