Question

Find the coordinates of the vertex for the parabola defined by the given quadratic function.

f(x)=5x^2-40x+22

Answer 1

well start by grouping things

\[f(x) = (5x^2 + 40x) + 22\]

remove the common factor from the brackets

\[f(x) = 5(x^2 + 8x ) + 22\]

you will need to complete the square for the expression in the brackets.

you goal is to get it to the form

\[y = a(x - h)^2 + k\]

so what gets added into the brackets to make a perfect score...?

Answer 2

thats one option....

the alternate option is

find the line of symmetry

for a parabola

\[y = ax^2 + bx + c\]

the line of symmetry is

\[x = \frac{-b}{2a}\]

in your question b = 40 and a = 5

substitute them to find x...

the vertex is on the line of symmetry... so substitute the x value you just found to get the y value of the vertex

Answer 3

Ok I'll probably go with your first way of solving it!

Answer 4

why did you only take 5 out instead of 5x?

Answer 5

well if you use the 2nd method

\[x = \frac{-40}{2 \times 5}\]

so

x = - 4

this is the x- part of the vertex

just substitute x = -4 to find the y part of the vertex..

Answer 6

is this right? For the first method...

5(x+4)^2+38

Answer 7

completing the square gave me (4+x)^2 than I added 16 to the 22

Answer 8

well almost

(5(x + 4)^2 is great....

it expands to

5x^2 + 40x + 80...

so to compensate you need to subtract 80

so you have

y = 5(x + 4)^2 + 22 - 80

that keeps it in balance... so simplify the last bit to get it into vertex form.

Answer 9

-4,-58?

Answer 10

yep... seems to make sense to me...

Answer 11

-4 isn't right unfortunately

Answer 12

I'm doing my homework and it says it's wrong.. but that -58 is right which doesn't make sense to me because I used -4 to get that

Answer 13

Oh it's 4

Answer 14

no its -4.... unless you original equation is entered incorrectly...

you can graph the curve using this site

https://www.desmos.com/calculator

and see the vertex is -4, -58

Answer 15

so there are a couple of options...

the equation should be

y = 5x^2 - 40x + 22

or... there is an error in entering the answer..... someone forgot to enter the negative when the answer was created.

Answer 16

Maybe that was the case. So for this one, 2x^2-28x-144 I used your symmetry equation and got x=7

Answer 17

then I plugged that in to get -242, but those are wrong. What did I do wrong?

Answer 18

It is also asking for the x and y intercepts of that line

Answer 19

so x = 7 is correct

so its 2(7)^2 - 28(7) - 144 = -242

thats correct as well

well the y- intercept is y = -144

and the x intercepts are from factoring

so

y = 2(x^2 - 14x - 72)

y = 2(x -18)(x + 4)

so the x intercepts are 18, -4

are you going from a text or website...?

because of you use something like wolframalpha

http://www.wolframalpha.com/

it will get the vertex

which is the same as the ones you have found

Answer 20

I'm confident, based on the 2 equations you posted, the vertex for each equation is correct.

I'd question your teacher...

Answer 21

how do i get the y intercepts?

Answer 22

y- intercepts are easy.... just let x = 0... and then you get y = -144

Answer 23

The part I really struggle with is finding the domain and range for equations like this. How do I do that?

Answer 24

ok... so the domain.... what values of x can you put into the equation..

in both parabola... you can enter any value of x.... so the domain is all real x.

the range is what are the output values .....

in both these parabolas they are concave up.... and the lowest value is the y part of the vertex....

so the range is y > vertex value

the 2nd question should have a range y >= -242

Answer 25

I have to go to school... hope it helped...

your answers are correct...

Answer 26

thank you for your help! Take care!