SOLUTION: Describe the graph of the following quadratic equations; include the x-intercepts, y-intercept, two other points on the parabola, equation of the line of symmetry, vertex and tell

Question 42881: Describe the graph of the following quadratic equations; include the x-intercepts, y-intercept, two other points on the parabola, equation of the line of symmetry, vertex and tell whether the parabola opens up or down. Show ALL math used in each process.
a. y = x^2 - 3
b. y = x^2 + 3x - 4
c. y = -x^2 + 2x
Answer by ilana(307) (Show Source): You can put this solution on YOUR website!
I'll do part b since it is probably the most complicated.
If we have the parabola in the form y-k=a(x-h)^2, we know the vertex is (h,k), so the line of symmetry is x=h.
So for part b, we have y+4=x^2+3x. The biggest issue here is that x^2+3x does not fit the form (x-h)^2. So, we make it fit this form by "completing the square". So we need to add (3/2)^2 to the right side, so we add it to the left side as well. So we have y+4+9/4 = x^2+3x+9/4, so y+25/4 = (x+3/2)^2.
Now we have our parabola in a form we could work with.
So the vertex is (-3/2,-25/4) and the line of symmetry is x=-3/2. We get the x-intercept by solving for y=0 and the y-intercept by solving for x=0. 25/4=(x+3/2)^2 gives us x=1,-4 for the x-intercepts, and y=(3/2)^2-25/4 gives us y=-4 for the y-intercept. Notice that the 2 x-intercepts and 1 y-intercept makes sense since we have a "U" going up or down vertically. Since a in our equation, which happens to be 1, is positive, this parabola opens up, making it look like a regular "U". For 2 other points on the parabola, you can just plug in a random value for x and then solve for that y, or you can get two points at once by plugging in a value for y (it must be greater than the y-value of the vertex) and solving for the 2 corresponding values of x.
Good Luck!!
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