Question 29731: I need help graphing these functions could I get some help please?
y=4x^2
and graph function y=x^2-2
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! For the first function:  , you know (or should know) that this is the equation for a parabola that opens upwards (the coefficient of x^2 is positive) and the x-coordinate of the vertex is given by:  .
The a and b come from the standard form for the quadratic equation:  , so the vertex of the parabola is:  because b = 0, and, in your equation, when x = 0 then y = 0, so the vertex of the parabola is at (0, 0), the origin of the coordinate system.
The x-intercepts can be found by setting y = 0 and solving for x.
and or, there are no x-intercepts which makes sense when you realise that if the parabola opens upwards and the vertex is at the origin, then the parabola never crosses the x-axis.
Here's the graph:
For the second function, you can apply a similar analysis and find that the parabola opens upwards but the vertex is at (0, -2) and the x-intercepts are at x =  and x = -
Here's the graph:
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