SOLUTION: say you have the equation y = x^3 - 3. How would you solve the y-intercept, the x-intercept and the vertex to this equation?

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: say you have the equation y = x^3 - 3. How would you solve the y-intercept, the x-intercept and the vertex to this equation?      Log On


   

Question 11060: say you have the equation y = x^3 - 3. How would you solve the y-intercept, the x-intercept and the vertex to this equation?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The y-intercept for y+=+x%5E3+-+3 is found by setting x = 0.
y+=+%280%29%5E3+-+3
y+=+-3 The y-intercept is at (0, -3)
The vertex? This question would make sense for a quadratic equation, but you have a cubic (third degree) equation and these have no vertex (maximum or minimum). The best you can do for a cubic function is to find the relative maximum or relative minimum, if there is one.
Take a look at the grah of this function and you'll see what I mean.
graph%28300%2C200%2C-5%2C5%2C-5%2C5%2Cx%5E3-3%29
You can see that the y-intercept is (o, -3) but the vertex?
Perhaps you meant to type: y+=+x%5E2+-+3 ??
Let's see what that looks like.
graph%28300%2C200%2C-5%2C5%2C-5%2C5%2Cx%5E2-3%29
The y-intercept is found, as before, by setting x = 0.
y+=+%280%29%5E2+-+3
The y-intercept is at (0, -3) as you can see on the graph.
The x-coordinte of the vertex is found by:
x+=+-b%2F2a
x+=+-%280%29%2F2
x+=+0 and the y-coordinate of the vertex is found by substituting this value of x into the original equation and solving for y.
y+=+%280%29%5E2+-+3
y+=+-3
The y-coordinate is y = -3.
So the vertex is at (0, -3)