SOLUTION: The graph of y=(m-3)-mx-x^2 touches the x-axis I have calculated the first two parts of the question which give me: When m=2, x intercept =-1 When m=-6, x intercept =3 So my qu

Algebra ->  Finance -> SOLUTION: The graph of y=(m-3)-mx-x^2 touches the x-axis I have calculated the first two parts of the question which give me: When m=2, x intercept =-1 When m=-6, x intercept =3 So my qu      Log On


   

Question 909158: The graph of y=(m-3)-mx-x^2 touches the x-axis
I have calculated the first two parts of the question which give me:
When m=2, x intercept =-1
When m=-6, x intercept =3
So my question is: For each set of values found, find the area of the triangle formed by the x-intercepts and y-intercept.
Please help me with this question
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
when m = 2, we have y = -x^2 -2x -1
when m = -6, we have y = -x^2 +6x -9
the above are two equations of parabolas that curve downward
a) y = -x^2 -2x -1 factors into (-x-1) * (x+1) and
x intercept is -1
y intercept is -1
b) y = -x^2 +6x -9 factors into (-x+3) * (x-3) and
x intercept is 3
y intercept is -9
c) for each triangle the base is 4 units
Area (A) of triangle is (1/2)*base*height
for a) A = (1/2)*4*1 = 2
for b) A = (1/2)*4*9 = 18