SOLUTION: f(x)=1/2x^2+6x+7
I know that the vertex is (-6,-11)
I know that the graph opens up.
I know the equation of the axis of symmetry is -6.
What I need to figure out is wha
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Question 487788: f(x)=1/2x^2+6x+7
I know that the vertex is (-6,-11)
I know that the graph opens up.
I know the equation of the axis of symmetry is -6.
What I need to figure out is what the x intercepts are.
It says to type the exact answer using radicals as needed. Simply as needed. Do not factor.
Geee I am so confused!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
f(x)=1/2x^2+6x+7
I know that the vertex is (-6,-11)
I know that the graph opens up.
I know the equation of the axis of symmetry is -6.
What I need to figure out is what the x intercepts are.
It says to type the exact answer using radicals as needed. Simply as needed. Do not factor.
----------
x-intercepts:
Let y = 0, and solve for "x" using the Quadratic Formula:
1/2x^2+6x+7 = 0
a = (1/2) ; b = 6 ; c = 7
---
x = [-b +- sqrt(b^2-4ac)]/(2a)
----
x = [-6 +- sqrt(36 - 4*(1/2)*7)]/(1)
---
x = [-6 +- sqrt(12)]
---
x = -6+2sqrt(3) or x = -6-2sqrt(3)
=======================================
Cheers,
Stan H.
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