ʕ•ᴥ•ʔ Find the Equation of a Parabola from a Graph with an Easy Walkthrough

In the last lesson, we learned how to draw a parabola from its function. This lesson, we will do the opposite. We will learn how to find the quadratic function when we are given the graph of a parabola.

Here we are given a parabola with the vertex at negative 1 and 4. Vertex. The question explicitly tells us the information for the vertex, that means we should use the vertex form to model this parabola, right. Makes sense? Yeah. So let's use the vertex form to model this parabola. Okay? Let's go. So vertex form goes something like this, right, with y on the left side of the equation, x on the right side of the equation with the square right here meaning we are dealing with a quadratic function, right, cool. And the a here is the leading coefficient for the parabola, okay? Cool. Now, vertex. So this vertex information should be able to tell us the information about something here, and something here, right? So let's go. For this vertex the x coordinate is negative 1. So if you want to make the statement look something like this, with a bracket and the x inside, we are looking for a value here, right? Would you say that means we have to move the number to the same side as x, and we can put a bracket around the statement, right? So let's do this. Now, negative 1 is moved to the other side of the equation, it becomes positive 1, right? And now the rest of the equation we just have zero, and now we can put a bracket around this whole thing. And that is gonna be the information we are gonna put down here, right, yeah. So inside the bracket what do we see? X plus 1. So inside the bracket, we just put down x plus 1. And that's done, okay? So the x portion is done. Now let's take care of the y portion. So for the vertex, the y coordinate is 4. Let's do the same thing, move the number to the same side as y. So positive 4 moved to the other side becomes negative 4, and this side of the equation we just have zero. And now, we can put a bracket around this, right. And now we can put this whole thing right here. So inside the bracket we have y minus 4. Inside the bracket we have y minus 4. And now this portion of the equation is done. So now how do you determine the value for the leading coefficient? Well there's still one information we haven't used, so chances are this will help us find a leading coefficient: the information about this point. So let's go. What does this point tell us? What information can we get out of this point? Well, this point is basically telling us that for this quadratic function, when x equals negative 3, the corresponding y value is 12. And guess what? These other values will plug into the equation to find the leading coefficient. Okay? So here, x value, we're going to plug in negative 3. Here, the y value, we're gonna plug in 12. Okay guys? And let's try to solve for a. So left side of the equation we have 12 minus 4, we get-- let me use another color. So, left side of the equation we have 12 minus 4 we get 8 .equals, a is the value we are trying to find. Now, this side we have negative 3 plus 1, we have negative 2. And don't forget outside we gotta have a square, right. Cool. Now, a equals negative 2 squared. Guys, negative 2 squared we get positive, right. Positive of what, 2 squared is 4. So we have 4a. Okay guys. Now, to solve the value for a, to solve the value for a, we simply divide both sides of the equation by 4. That way 4, 4 cancels out. So at the end, a equals 8 divided by 4 is 2. That's it, that's the value for a right here. Okay, guys? So at the end, for this parabola, the quadratic function is right here, okay? Left side of the equation we have y minus 4 equals. . . Guys, what's the value for a right here? A is 2, right? So a is 2. So we put it in right here. Now bracket, inside the bracket we have x plus 1.

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[autumnreed]

If you’re given the roots instead you can use the formula y=a(x-root1)(x-root2)

For example, let’s say you’re given a parabola with the roots (where the line hits the x-axis) are x=-1 and x=3

Take the above formula, and plug in the roots where it says root1 and root2

y=a(x+1)(x-3)

You flip the signs for the same reason he explained in this video; x=-1 👉🏼 x+1=0

You then take another point on the graph and plug in the x and y values. Let’s say this other point is (0, 3)

3=a(0+1)(0-3) *Yes, you will be plugging in the same x value twice.

You then solve, and come up with:
-1=a

Now that you know a=-1, it’s time to plug it into the original equation.

y=-1(x+1)(x-3) *You will leave the roots in the equation. Do not add the x and y values of the other point you used prior. That was just in order to solve for a.

This is in factored form. If you want in standard form all you have to do is distribute, which I’ll show below just for the sake of it.

y=-1(x+1)(x-3)
y=-1(x^2 - 2x - 3) *You have to distribute what’s in both of the parenthesis first. x times x, x times -3, then 1 times x, and one times -3. Combine like terms.

Standard form: y=-x^2 + 2x + 3

Hope this helped. OH WAIT...

If it only has one root, just plug in the root two times. Then it’s the same process.

OK I’m done now.

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