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Find a quadratic function that has a y-intercept of y=4 and is increasing on the point (3,28).
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have you tried to solve this?
no i cant do it
i think its f(x)=x^2-4
no, if you let x=0 which is the y intercept then you get y=-4 let me assist
the quadratic has to be on the form y=ax^2+bx+c when x=0 the y intercept is c so what do you think c should be?
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i dont know
just substitute x=0 into the general quadratic equn and you will get a y value
and you already know that the y-intercept has to be 4 i.e. when x=0, y=4
so what should c equal?
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