Question 62769: can you help?
find the vertex and intercepts of the quadratic function f(x)=x^2+x-6. thank you very much
Found 2 solutions by uma, jai_kos:
Answer by uma(370)   (Show Source):
You can put this solution on YOUR website! The given equation is y = x^2 + x - 6
This is of the form ax^2 + bx + c = y
The x co-ordinate of the vertex = -b/2a
= -1/2*1
= -1/2
Replacing x by -1/2 in the given equation,
y = (-1/2)^2 + (-1/2) - 6
= 1/4 - 1/2 - 6
= -1/4 - 6
= - 6 1/4
The vertex = (-1/2, - 25/4)
To find the y intercept we replace x by 0.
==> y = 0 + 0 -6
==> y = - 6
To find the x intercept we set y = 0
==> x^2 + x - 6 = 0
==> x^2 + 3x - 2x - 6 = 0
==> x(x+3) - 2(x+3) = 0
==> (x+3)(x-2)= 0
==> x+ 3 = 0 or x - 2 = 0
==> x = - 3 or x = 2
Thus the x intercepts are -3 and 2 while the y intercept is -6
Good Luck!!!
Answer by jai_kos(139) (Show Source):
You can put this solution on YOUR website! f(x)=x^2 + x- 6
vertex and intercepts of the quadratic function
Given f(x) = x^2 + x- 6 --->(1)
It is a quadratic equation, where a = 1, b =1 and c = -6
vertex = - b /2a = -1 /2*1 = - 1/2
x = -1/2
Now substitute for x = -1/2,
Put this value in equation(1), we get
f(-1/2) = (1-/2)^2 + (1/2) - 6
= 1/4 + 1/2 - 6
= 3 /4 - 6
= -21 /4
Therefore the vertex is given by,
(x, f(x) = ( -1/2, -21/4)
To find the intercepts, we put x =0 in equation(1) we get
y = f(0) = -6
THis is the y-intercepts
Now put f(x) = 0 in the equation(1), we get
x^2 +x - 6 = 0
x^2 +3x -2x -6 = 0
x(x+3) -2(x+3) = 0
(X-2)(x+3)= 0
Therefore x = 2 or -3
The x intercepts are x =2 and x =-3
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