SOLUTION: which parabola touches the x-axis at one point only? a)y=x^2+8x+16 b)y=x^2-16 c)y=x^2-5x+6 d)y=x^2+4

Question 77234: which parabola touches the x-axis at one point only?
a)y=x^2+8x+16
b)y=x^2-16
c)y=x^2-5x+6
d)y=x^2+4
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
If we complete the square for the quadratic, we can find the vertex of the quadratic.

Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form

Start with the given equation

Subtract from both sides

Factor out the leading coefficient

Take half of the x coefficient to get (ie ).

Now square to get (ie )

Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of does not change the equation

Now factor to get

Distribute

Multiply

Now add to both sides to isolate y

Combine like terms

Now the quadratic is in vertex form where , , and . Remember (h,k) is the vertex and "a" is the stretch/compression factor.

Check:

Notice if we graph the original equation we get:

Graph of . Notice how the vertex is (,).

Notice if we graph the final equation we get:

Graph of . Notice how the vertex is also (,).

So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.

So using the standard vertex form:

where a=1, h=-4, and k=0
So the vertex is (-4,0) which shows that it touches at one point. It turns out that any perfect trinomial will have only have one x-intercept.
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