Question 178086: Quadratic Equations
Find the x-intercepts of each parabols.
c) y=x^2-4x+4
d) y=4x^2-12x+9
Thank you very much pleaseeeeeeeeeeee
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Quadratic Equations
Find the x-intercepts of each parabola.
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The x-intercepts are the points that y = 0, so set the function = 0 and solve.
c) y=x^2-4x+4
x^2 - 4x + 4 = 0
(x-2)*(x-2) = 0
Only one intercept, at (2,0)
| Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=0 is zero! That means that there is only one solution:  .
Expression can be factored: 
Again, the answer is: 2, 2.
Here's your graph:
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d) y=4x^2-12x+9
4x^2 - 12x + 9 = 0
(2x-3)*(2x-3) = 0
Same here, one point where the parabola is tangent to the x-axis.
| Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
Quadratic equation (in our case  ) has the following solutons:
For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=0 is zero! That means that there is only one solution:  .
Expression can be factored: 
Again, the answer is: 1.5, 1.5.
Here's your graph:
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