Question

axis parallel to the y axis , passes through (-3,2),(0,-5/2) and (1,-6)

Answer 1

that is a parabola sir

Answer 2

@thormaster

Answer 3

@thomaster

Answer 4

@tetsuguy
1/am I right?
2/ work on it or wait for thomaster?

Answer 5

what sir ?? @Loser66

Answer 6

we are supposed to write the equation of the parabola, right?

Answer 7

yes sir

Answer 8

sir

Answer 9

since the parabola has symmetry // to y -axis, it has the form of y = ax^2 +bx +c
All you have to do is plug the points into the form to find out a, b, c
I do the first point for you as example, you just imitate the steps to solve
for the point (-3,2) , that means x =-3, y =2
Plug into the form
\[\color{red}{y}=a\color{red}{x^2}+b\color{red}{x}+c\\\color{red}{2}=a\color{red}{(-3)^2}+b\color{red}{(-3)}+c\\2=9a-3b+c (*)\]
you do the same with the other points, so you will have 3 equations, 3 unknowns a, b, c
Solve that system, you have the answer

Answer 10

i dont know your formula sir :(

Answer 11

If you don't know the formula of the parabola, how can we step up?
I am sorry, I waste your time. Let me compensate by tagging others. Hopefully they can help
@mathmale

Answer 12

the formula of an parabola are (y-k)sqrd.=+-4p(x-h) and for parallel to 0y is (x-h)sqrd.+-4p(y-k)

Answer 13

oh, I see, the vertex-directrix form

Answer 14

ok, if you are familiar with that form, work on it,!! the process is the same. Plug the points into the form and solve for h,k , p

Answer 15

:/ how can i

Answer 16

Any (correct) formula for a quadratic function can be re-written in any other form desired. Personally I'd go with Loser66's \[y=ax^2+bx+c\]
This form of the equation of a quadratic occurs very frequently, and so should be on your "must know" list of basic functions.

Answer 17

Wait for professor @mathmale

Answer 18

Supposing we do start out with \[y=ax^2+bx+c\] and that we take one of the three given points, (0,-3/2) (which happens to be the y-intercept). Here, when x=0, y=-3/2. Substituting these values into \[y=ax^2+bx+c\]

Answer 19

we get\[-3/2 = y=a(0)^2+b(0)+c\]

Answer 20

which immediately tells us that c=-3/2.

Answer 21

Now re-write the quadratic using that value of c:\[y=ax^2+bx-3/2\]
and similarly substitute the coordinates of each of the two remaining given points to find the values of a and b.

Answer 22

Consider the point (-3,2): x=-3 and y=2. Substitute these values into \[y=ax^2+bx-3/2\]now, please.

Answer 23

Please show your work. You'll get an expression that looks like this:\[( )=a( )^2+b( ) - 3/2\]