My daughter is working on algebra that is a bit difficult. I have not been able to figure out how to help her with the graphs, the velocitiy, etc. Here is an example of one of the questions - y=2X^2, -3. Can you explain how to figure out this type of equation?

I agree with Brian that this is a quadratic function and will look like a parabola.

If I am reading the equation correctly, the problem is the following:

y = 2x^{2} - 3

Quadratic equations are in the form of y = ax^{2} + bx + c, and although there are several ways to solve quadratic equations, the old standard is the quadratic formula:

x = (-b +- sqrt(b^{2}-4ac))/2a

In your equation, there is no x value so the value of b is 0.

If there are two solutions to the equation (from the 'plus or minus' sign), these x values represent where the parabola crosses the x-axis on a graph. The -3 in your equation represents where the parabola crosses the y-axis. The vertex (maximum or minimum) of the parabola is determined from -b/2a.

To be completely accurate, the part of the equation (called the determinant) that is b^{2}-4ac may be > 0, = 0, or < 0.

If the determinant is positive, then there will be two values for x.

If the determinant equals 0, there will only be one value of x.

If the determinant is negative, there are no real solutions to the equation. This is an advanced concept for late algebra 2 or calculus.

To create any graph, you need to select a set of x values, substitute, and solve the equation for y so you have a set of points to plot. For this equation,

when x = 0, y = -3

Similarly x = 1. y = -1

x = -1, y = -1

x = 2, y = 5

x = -2, y = 5

Plotting these points then sketching a curve through them should give you an idea of what the parabola looks like.

I hope this helps.

Lynne

## Comments

Dear Lela, help us to help you, copy the question exactly as it appear in the original source and separate it from your personal comments. Otherwise there will be "may be... might be... could be..." in our answers.