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Find the minimum y-value on the graph of y=f(x). f(x)=x^2+8x-6 the minimum y-value is y=
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Well it's a parabola with a positive a value so the minimum y-value is at the vertex. x-coordinate of vertex=-b/2a. to find y-coordinate, plug in the x-coordinate of the vertex into the original equation.
can you help me with that
Alright. Well parabolas are in the form y=ax^2+bx+c. So in this case, a=1, b=8 and c=-6 (make sure you understand how I got these numbers). x of vertex = -b/2a= -8/(2*1)=-8/2=-4 y of vertex = (-4)^2+8(4)-6=16+32-6=42
ok now how do I find the y- value
-_- it's in my post
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would it be the y of vertex
is the y value the the yof vertex
In my first post: "minimum y-value is at the vertex"
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