What is the value of h when the function is converted to vertex form?

note: vertex form is f(x)=a(x−h)2+k .

f(x)=x2+10x+35

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h =

what is the minimum value for h(x)=x2−16x+60?

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y =

what are the x-intercepts of the quadratic function?

f(x)=x2−3x−10

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let f(x)=x2+17x+72 .

what are the zeros of the function?

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and

let f(x)=x2−8x+19 .

what is the minimum value of the function?

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Rewrite as perfect squares

f(x)=(x+5)²+10

(h,k) is (-5,10)

the answer Part 1) is 

h is -5

Part 2) What is the minimum value for h(x)=x²−16x+60?

h(x)=x²−16x+60

Group terms that contain the same variable

h(x)=(x²−16x)+60

Complete the square . Remember to balance the equation

h(x)=(x²−16x+64)+60 -64

Rewrite as perfect squares

h(x)=(x-8)²-4

(h,k) is the vertex> (8,-4)

the answer Part 2) is

the minimum value of h(x) is -4

Part 3)

Group terms that contain the same variable, and move the constant to the opposite side of the equation

Rewrite as perfect squares

Group terms that contain the same variable, and move the constant to the opposite side of the equation

Rewrite as perfect squares

(x+8.5)²=0.25> (+/-)[x+8.5]=0.5

(+)[x+8.5]=0.5> x1=-8

(-)[x+8.5]=0.5> x2=-9

the answer part 4) is 

x=-8

x=-9

Rewrite as perfect squares

f(x)=(x-4)²+3

the vertex is the point (4,3)

the answer Part 5) is 

the minimum value of the function is 3 

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490

which expression represents the approximate length of bc?

step-by-step explanation: