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Mathematics
OpenStudy (anonymous):

Write an equation for the parabola with a vertex at (2, -1) and y-intercept 5.

OpenStudy (saifoo.khan):

\[y = (x - 2)^2 -1\]

OpenStudy (amistre64):

when x=0 that equals 3 tho, not 5

OpenStudy (saifoo.khan):

Messed up.. o_O

OpenStudy (saifoo.khan):

Yeah..

OpenStudy (saifoo.khan):

figured//

OpenStudy (amistre64):

gotta scale it right?

OpenStudy (amistre64):

5 = a(-2)^2-1

OpenStudy (amistre64):

its a good start tho, get the basic down and fidget with it

OpenStudy (anonymous):

That was what the question asked, I am confused on how to calculate (a)

OpenStudy (amistre64):

a is just a single unknown that can be algebraed into submission

OpenStudy (amistre64):

add 1 and divide off the (-2)^2

OpenStudy (anonymous):

My algebra teacher said the answer was \[3/2(x-2)^{2}\]

OpenStudy (anonymous):

PLus 1 at the end sorry

OpenStudy (amistre64):

almost, missing the -1 at the end tho

OpenStudy (anonymous):

oh minus?

OpenStudy (anonymous):

Ok so how did he get 3 halves?

OpenStudy (amistre64):

5 = a(-2)^2 - 1 +1 +1 ---------------- 6 = a(4) /4 /4 --------- 6/4 = a

OpenStudy (amistre64):

this is basic equation manipulations; you will need to become quite adept at it for further math skills

OpenStudy (anonymous):

You put in the y intercept (5) in for x?

OpenStudy (amistre64):

when x=0, y=5 so we zero out the x and replace y by 5

OpenStudy (amistre64):

y = a(x-2)^2 - 1 5 = a(-2)^2 - 1

OpenStudy (anonymous):

Is it also possible to substitute h and k as well instead?

OpenStudy (amistre64):

h and k are your vertex elements

OpenStudy (amistre64):

we plugged those in to determine "a"

OpenStudy (anonymous):

oh ok i get it (hopefully) thank you

OpenStudy (amistre64):

practice helps :)

OpenStudy (saifoo.khan):

Practise makes a man perfect. ;)

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