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

Algebra ->  Rational-functions -> 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      Log On


   

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) About Me  (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


y=1+x%5E2%2B8+x%2B16 Start with the given equation



y-16=1+x%5E2%2B8+x Subtract 16 from both sides



y-16=1%28x%5E2%2B8x%29 Factor out the leading coefficient 1



Take half of the x coefficient 8 to get 4 (ie %281%2F2%29%288%29=4).


Now square 4 to get 16 (ie %284%29%5E2=%284%29%284%29=16)





y-16=1%28x%5E2%2B8x%2B16-16%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 16 does not change the equation




y-16=1%28%28x%2B4%29%5E2-16%29 Now factor x%5E2%2B8x%2B16 to get %28x%2B4%29%5E2



y-16=1%28x%2B4%29%5E2-1%2816%29 Distribute



y-16=1%28x%2B4%29%5E2-16 Multiply



y=1%28x%2B4%29%5E2-16%2B16 Now add 16 to both sides to isolate y



y=1%28x%2B4%29%5E2%2B0 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=-4, and k=0. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2%2B8x%2B16 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2%2B8x%2B16%29 Graph of y=1x%5E2%2B8x%2B16. Notice how the vertex is (-4,0).



Notice if we graph the final equation y=1%28x%2B4%29%5E2%2B0 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x%2B4%29%5E2%2B0%29 Graph of y=1%28x%2B4%29%5E2%2B0. Notice how the vertex is also (-4,0).



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:
y=a%28x-h%29%5E2%2Bk
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.