SOLUTION: Without drawing the graph determine how many x- intercepts the parabola has and whether its vertex lies above or below or on the x axis Show your worky y = -x^2 + 2x - 1

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Without drawing the graph determine how many x- intercepts the parabola has and whether its vertex lies above or below or on the x axis Show your worky y = -x^2 + 2x - 1      Log On


   

Question 203135: Without drawing the graph determine how many x- intercepts the parabola has and whether its vertex lies above or below or on the x axis Show your worky
y = -x^2 + 2x - 1
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
X-intercepts are where a a graph intersects the x-axis. In other words, x-intercepts are x-values which make the function value 0. So to find x-intercepts we need to find the x-value(s) where f(x) = 0.

So finding the x-intercepts of your function means solving:
0+=+-x%5E2+%2B+2x+-+1
Since solving quadratic equations by factoring is easier than using the quadratic formula and since factoring is easier when the leading (first) coefficient is 1, I am going to start by factoring out a -1:
0+=+-1%28x%5E2+-+2x+%2B+1%29
This factors easily:
0+=+-1%28x+-+1%29%5E2
The only way for a product to be zero is if one of the factors is zero. The only factor that could be zero is (x-1). So the only solution to 0+=+-1%28x+-+1%29%5E2 is the solution to (x - 1) = 0, which is x = 1. So there is only one x-intercept: (1, 0)

As for where the vertex is located, think about parabolas in general. In general the possibilities are:
  • 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.
  • Part of the parabola is above the x-axis and part of it is below the x-axis. These parabolas have 2 x-intercepts. If the leading coefficient is positive the vertex will be below the x-axis. If the leading coefficient is negative, the vertex will be above the x-axis.
  • Parabolas whose vertex is not above or below but on the x-axis. These parabolas will have just one x-intercept, the vertex.

Our parabola, with just one x-intercept, is in the last category. Its vertex is not above or below the x-axis. It is on the x-axis. It is the x-intercept: (1, 0).