Question

Suppose the function shown below was expressed in standard form,y = ax^2 + bx + c . What is the value of a?
Graph - http://i.imgur.com/R4uP5NA.png

Answer 1

Well, you can see what the roots are, from the x-intercepts. So that allows you to write an equation of the form:

y=(x - a)(x - b) where a and b are the two roots (which are the x-intercepts).

Now the only question is, is that the equation or do you need to apply a vertical stretch or shrink? So once you have that equation (and multiply it out), check the value you get for the vertex.

The x-coordinate of the vertex is x=-b/(2a), then evaluate the function for that value of x. If what you get is the y-coordinate of the vertex, then a=1. If not, then choose a so that you get the vertex coordinate as pictured on the graph.

Answer 2

so it as to be 1 or -1, i'm a bit confused let me read over what you said again

Answer 3

sorry debbie, I just don't get it. Is there anyway you can explain it to me in a more simple way. :/

Answer 4

You have x-intercepts at x=0 and x=2, right?

That means that (x - 0) and (x - 2) are factors of the equation, e.g.:

y=a*x*(x-2) , so \(y=a(x^2-2x)\)

So the graph is some multiple (which means a vertical stretch or shrink) of:

\(y=x^2-2x\)

Now, MAYBE that's the function - if so, that would mean that a=1. Maybe it isn't. To find out, find the vertex of \(y=x^2-2x\), using the formula for the vertex that x=-b/2a, and y=f(-b/2a). (Tell me of you understand that formula, and how you use it to find the vertex).

You WILL get the correct x-coordinate of x=1, because a vertical stretch doesn't affect the x-coordinate of the vertex (and the x-coordinate of the vertex is ALWAYS midway between the x-intercepts).

The question is, do you get the correct y-coordinate, which should be y=-2 (do you see that on the graph?).

Your vertex needs to be (1, -2). So if \(y=x^2-2x\) DOESN'T have the required y-coordinate for the vertex, then you need to multiply the whole equation by whatever value of a it takes, to get y=-2 when x=1.