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 work y = x^2 - 5x + 6

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> 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 work y = x^2 - 5x + 6      Log On


   

Question 203134: 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 work
y = x^2 - 5x + 6
Found 2 solutions by jsmallt9, jojo14344:
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+-+5x+%2B+6
Factoring this we get:
0+=+%28x+-+2%29%28x+-+3%29
The only way for a product to be zero is if one of the factors is zero. So the solution to 0+=+%28x+-+2%29%28x+-+3%29 is: %28x+-2%29+=+0 or %28x+-+3%29+=+0. The solutions to these are: x = 2 or x = 3.
There are two x-intercepts: (2, 0) and (3, 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 has 2 x-intercepts and its leading coefficient is 1 (which is positive). So our vertex is below the x-axis.

Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!


Given: system%28y+=+x%5E2+-+5x+%2B+6%29, wheresystem%28a=1%2Cb=-5%2Cc=6%29}
Finding for X-intercepts via Quadratic:
x%5E2-5x%2B6=0, a perfect square
%28x-3%29%28x-2%29=0
red%28x=3%29 & red%28x=2%29
X-intercepts --> (3,0) & (2,0) Answer

For the vertex:
x=-b%2F2a=-%28-5%29%2F%282%2A1%29=red%285%2F2%29 ---> substitute to the eqn,

y=%285%2F2%29%5E2-5%285%2F2%29%2B6=25%2F4-25%2F2%2B6
y=%2825-50%2B24%29%2F4=red%28-1%2F4%29

The vertex: (5/2,-1/4) or (2.5,-.25) ---> Lies below the X - axis being the Y-Intercept is negative (-).
To clearly demonstrate this, we see the graph


Thank you,
Jojo