MHB Quadratics: How to determine parabola equation.

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Hi all. I need a bit of help determining the equation of some parabolas given points, intercepts and vertexes. Below are the exacts questions, any help will be much appreciated as I need this done soon!

1. A parabola has turning point (1,6) and passes through the point (-1,8). Find its equation.

2. A parabola has x-intercepts of 2 and 7 and passes through the point (3,6). Find
the equation of the parabola.

Please show working out!
 
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James400 said:
Hi all. I need a bit of help determining the equation of some parabolas given points, intercepts and vertexes. Below are the exacts questions, any help will be much appreciated as I need this done soon!

1. A parabola has turning point (1,6) and passes through the point (-1,8). Find its equation.

2. A parabola has x-intercepts of 2 and 7 and passes through the point (3,6). Find
the equation of the parabola.

Please show working out!

My reply to your other post holds here too. Please consider posting some of your work, or letting us know what you don't understand.
 
I'm a little confused as to where to start as a matter of fact!:) I know the final objective is to reach the equation in standard form using the information provided, but I'm not sure how to do that (the processes involved). I was just hoping to get a step through of the processes required to reach the answer.
 
James400 said:
Hi all. I need a bit of help determining the equation of some parabolas given points, intercepts and vertexes. Below are the exacts questions, any help will be much appreciated as I need this done soon!

1. A parabola has turning point (1,6) and passes through the point (-1,8). Find its equation.

Since we are given the vertex, let's use the vertex form of a parabola:

$$y=a(x-h)^2+k$$

Since the vertex (turning point) is given as $(1,6)$, we may write:

$$y=a(x-1)^2+6$$

Now, to determine $a$, we may use the other given point $(-1,8)$ as follows:

$$y(-1)=a(-1-1)^2+6=8$$

Solving the above equation for $a$, what do you get?
 
MarkFL said:
Since we are given the vertex, let's use the vertex form of a parabola:

$$y=a(x-h)^2+k$$

Since the vertex (turning point) is given as $(1,6)$, we may write:

$$y=a(x-1)^2+6$$

Now, to determine $a$, we may use the other given point $(-1,8)$ as follows:

$$y(-1)=a(-1-1)^2+6=8$$

Solving the above equation for $a$, what do you get?

I'm all good with these questions now. I solved them a little while ago, but I'm still stuck on the other questions I posted.
 
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