Question

Check my answer:Without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis.

y = x^2 – 12x + 12


1 point in common; vertex on x-axis
2 points in common; vertex below x-axis
2 points in common; vertex above x-axis
no points in common; vertex above x-axis

Answer 1

no points?

Answer 2

Points in common with x-axis is same as where y=0,
which is = solving the quadratic equation for x:
x^2 - 12x +12 = 0
Vertex is where dy/dx = 2x -12=0

Answer 3

\(Hi\) \[y = x^2 – 12x + 12 \]
\[0 = x^2 – 12x + 12 \] Factor

But wait... this Parabola isn't factorable.

• The parabola is completely above the x-axis. These parabolas will have no x-intercepts and the leading coefficient is positive.

• The parabola is completely below the x-axis. These parabolas will have no x-intercepts and the leading coefficient is negative.

• The Parabola has no points if it is not factorable.

Answer 4

I notice that to choose appropriately from the four possible answers, we need to determine where the vertex of this parabola is, i. e., on, below or above the x-axis.

"Completing the square" would be an excellent tool for re-writing the given equation of this parabola in such a way that the vertex location is obvious. Are you familiar with "completing the square?"

If you carry out this operation, your result will look like this: y=a(x-h)^2 + k; (h, k) represents the coordinates of the vertex.

Answer 5

\[
b^2-4 a c==144-48>0

\]
The parabola cuts the x-axis in two different points

Answer 6

You can now check by graphing
http://www.wolframalpha.com/input/?i=plot+x%5E2+-+12+x+%2B+12

Answer 7

Since a > 0 the parabola opens up. Hence the vertex must be below the x-axis

Answer 8

Your second choice is the answer
2 points in common; vertex below x-axis

Answer 9

thank you guys!!

Answer 10

YW