SOLUTION: y= -1/2x^2 + 2x + 10
I have the vertex (2, 12) and the y intercept (0,10) but I keep getting stuck with....
x= -b +/- the square root of b^2 - 4ac over 2a.
I have it down t
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Question 272078: y= -1/2x^2 + 2x + 10
I have the vertex (2, 12) and the y intercept (0,10) but I keep getting stuck with....
x= -b +/- the square root of b^2 - 4ac over 2a.
I have it down to x= -2+/- the square root of 24 over -1 from there I am stuck
Answer by dabanfield(803) (Show Source): You can put this solution on YOUR website!
y= -1/2x^2 + 2x + 10
I have the vertex (2, 12) and the y intercept (0,10) but I keep getting stuck with....
x= -b +/- the square root of b^2 - 4ac over 2a.
I have it down to x= -2+/- the square root of 24 over -1 from there I am stuck
If you are supposed to solve the above for x we need to get the equation in the form:
ax^2 + bx + c = 0
-1/2x^2 + 2x + 10 = 0
In this case we have a = -1/2, b = 2 and c = 10. Using the quadratic formula then gives you:
-2 +- [sqrt(2^2 - 4*-1/2*10)]/(2*-1/2) =
-2 +- sqrt(4 + 20)/-1 =
-2 +- sqrt(24)/-1
The two solutions then are:
-2 - sqrt(24) and
-2 + sqrt(24)
sqrt(24) = sqrt(4*6) = sqrt(4)*sqrt(6) = 2*sqrt(6) so
we can rewrite the solutions as:
-2 - 2*sqrt(6)
-2 + 2*sqrt(6)
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