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Sonbull [250]
3 years ago
11

What is the vertex of the quadratic function f(x)=(x-8)(x-2)?

Mathematics
1 answer:
7nadin3 [17]3 years ago
4 0
Change the function into vertex form:
f(x)=x²-10x+16
make a square by (adding half of -10)², which is 25, then subtract it:
f(x)=x²-10x+25-25+16
f(x)=(x-5)²-9
so the vertex is at (5,-9)

another way to do it: 
f(x)=x²-10x+16
the vertex is when x=-\frac{b}{2a}, b=-10, a=1 in this case.
x=-(-10/2*1)=5
plug x=5 into the quadratic function, you get f(x)=-9
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Answer:

<h2> 22.2</h2>

Step-by-step explanation:

Step one

given the coordinates

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AB= \sqrt (-1+2)^2+(3+2)^2\\\\AB= \sqrt 1^2+(5)^2\\\\AB= \sqrt26\\\\AB=5.1

The distance between points BC=

BC= \sqrt (5+1)^2+(3-3)^2\\\\BC= \sqrt 6^2+(0)^2\\\\BC= \sqrt36\\\\BC=6

The distance between points CD

CD= \sqrt (4-5)^2+(-2-3)^2\\\\CD= \sqrt -1^2+(-5)^2\\\\CD= \sqrt26\\\\CD=5.1

The distance between points DA

DA= \sqrt (4+2)^2+(-2+2)^2\\\\DA= \sqrt 6^2+(0)^2\\\\DA= \sqrt36\\\\DA=6

Hence the perimeter = 5.1+6+5.1+6

= 22.2

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