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how would I find the vertex and x intercepts for f(x)= 25-x^2
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you can find the vertex by plugging into the equation -b/2a
so a=-1 and b=25
is c the y-intercept
for a parabola of that type, that is, with a squared "x" variable, you can get the vertex of it at \(\bf \left(-\cfrac{b}{2a}\quad, \quad-\cfrac{b^2}{4a}\right)\)
\(\bf \left(-\cfrac{b}{2a} \quad , \quad c-\cfrac{b^2}{4a}\right)\) rather
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