Question

One more please! Anyone wanna take a look? It's probably easier for you guys than me. Screenshot attached.

Answer 1

a=1, y=x^2 is not reflected or streched

Answer 2

You have an equation of the form ax^2 + bx + c = y where you can find 3 convenient points from the graph and plug them into the equation for 3 equations. I would use the y -intercept and the 2 x-intercepts. you will then have 3 equations in 3 unknowns (a,b,c) which will be very easy to solve.

Answer 3

You're not there yet, but I'll try to give you more here.

Answer 4

Try using points (-3,0), (1,0), and (0,-3) for starters.

Answer 5

This is where the graph hits the x and y axes.

Answer 6

No problem. The first equation is a(-3)^2 + b(-3) + c = 0. This comes from substituting in the first point into ax^2 + bx + c = y.

Answer 7

No, I could give you the answer to a, but I'm not allowed to just give an answer. I have to show you how to get there.

Answer 8

Try setting up the second equation after the pattern of what I did for you for the first.

Answer 9

That first equation simplifies to 9a - 3b + c = 0

Answer 10

If you don't like this approach, there is another "intuitive" way to get the answer and that is by visualizing shifting the graph so that the vertex is at the origin (the origin is the point (0,0) ).

Answer 11

You have to be good at visualizing because you will have to shift ALL the points (the whole graph) and see that you are dealing with a parabola.

Answer 12

Try that now and see if you can visualize what value of y you get for the new value of x.

Answer 13

Hint: shift the graph and take the new value of 1 for x. Get the y. Use the new x=2 and get the y. If you can see it, you will intuit the value for a. A third way is similar to the graph shift and is called a translation of axis but algebraically, that's the most involved, so you don't really have to go that route.

Answer 14

Are you getting this?

Answer 15

That good, but I want you to be able to see the problem to the end. Are you able to set up the second equation?

Answer 16

Don't beat yourself up, I'll help in any way I can, I'm very understanding.

Answer 17

That second point is (1,0) where x=1 and y=0. So, if we use ax^2 + bx + c = y, then we get a(1)^2 + b(1) + c = 0 or a + b + c = 0.

Answer 18

The 3rd equation is a(0)^2 + b(0) + c = -3 or c = -3. So, with c = -3, we already know one of the values for a,b, and c.

Answer 19

So, now you only have to solve the first 2 equations which are 9a - 3b = 3 and a + b = 3. If you solve for a and b, you will have your a.

Answer 20

This is merely solving 2 linear equations in 2 unknowns (a and b). Are you able to do that?

Answer 21

It's ok if you don't know how to do this, I can show you how and then you can derive the answer.