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Find the vertex, the line of symmetry the maximum or minimum value of the quadratic function, and graph the function f(x)=10-x^2
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okay, let's rearrange this. -x^2+10. so in the form ax^2 + bx + c, a is -1 b is 0 and c is 10.
axis of symmetry is -b/2a, and that is zero. Now, to find the vertex, plug in x=0 and you get y=10. So the vertex is (0,10).
that's also the minimum value, btw.
good?
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