Quadratics: How to determine parabola equation.

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In summary, In answer to question 1, the equation of a parabola is y=-a(x-h)^2+k. In answer to question 2, the equation of a parabola is y=a(x-h)^2+k.
  • #1
James400
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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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  • #2
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.
 
  • #3
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.
 
  • #4
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:

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

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

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

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

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

Solving the above equation for $a$, what do you get?
 
  • #5
MarkFL said:
Since we are given the vertex, let's use the vertex form of a parabola:

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

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

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

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

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

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.
 

1. What is a parabola equation?

A parabola equation is a quadratic function that represents the shape of a parabola. It is in the form of y = ax^2 + bx + c, where a, b, and c are constants and x is the independent variable.

2. How do you determine the vertex of a parabola?

The vertex of a parabola can be determined by using the formula (-b/2a, f(-b/2a)), where a and b are the coefficients of the quadratic equation in standard form. The vertex represents the highest or lowest point on the parabola.

3. What is the axis of symmetry of a parabola?

The axis of symmetry of a parabola is a vertical line that divides the parabola into two equal halves. It passes through the vertex of the parabola and is perpendicular to the directrix.

4. How do you find the x-intercepts of a parabola?

The x-intercepts of a parabola can be found by setting the equation equal to 0 and solving for x. This can also be done by factoring the quadratic equation or by using the quadratic formula.

5. How can you determine the direction of opening for a parabola?

The direction of opening for a parabola can be determined by looking at the coefficient of the x^2 term. If the coefficient is positive, the parabola opens upwards, and if it is negative, the parabola opens downwards.

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