SOLUTION: how do i solve for a vertex? when it asks for a Vertex of (0,0) and it is mimnimum?

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Question 1046925: how do i solve for a vertex? when it asks for a Vertex of (0,0) and it is mimnimum?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Depending how you have the equation, general form y=ax^2+bx+c or standard form y=a(x-h)^2+k; a%3E0 will make vertex to be a minimum. Vertex of (0,0) will make the standard form equation take the format y=ax^2. The vertex would be the point, (h,k).


y=x%5E2 is a reference equation for a parabola. Vertex will be the point (0,0). Setup a data table to find some point, graph the points, and sketch this and you will find a graph like this:
graph%28400%2C400%2C-10%2C10%2C-10%2C10%2Cx%5E2%29

If you shift the graph leftward or rightward, using subtraction of some value h, then in standard form, you would have an equation like y=%28x-h%29%5E2. This would be a shift TO THE RIGHT, IF h%3E0; or a SHIFT TO THE LEFT IF h%3C0. This means that the vertex, still touching the x-axis, will be at (h,0).
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Imagine that h=5.
The equation could become y=%28x-5%29%5E2 and the graph is this:
graph%28400%2C400%2C-10%2C10%2C-10%2C10%2C%28x-5%29%5E2%29


Imagine that h=-2.
The equation could become y=%28x-%28-2%29%29%5E2, or y=%28x%2B2%29%5E2 and this shifts the model reference graph two units LEFTWARD, still with vertex touching the x-axis.
The graph:
graph%28400%2C400%2C-10%2C10%2C-10%2C10%2C%28x%2B2%29%5E2%29.
and vertex is at (-2,0).


Looking again at standard form, y=%28x-h%29%5E2%2Bk, and not yet discussing how k value will contribute to the shifting of the graph, vertex is (h,k); and if the vertex is (0,0), the Origin, then putting these coordinates values into the standard form model, y=%28x-0%29%5E2%2B0, which simplifies to y=x%5E2.

I have used in these examples, a=1.