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well the vertex form of a parabola is \[y=a(x - h)^2 + k\] where (h, k) is the vertex. so substitute you vertex values, h = 4 and k = -1 into the equation. tThe y-intercept is on the parabola... so to find the value of a you need to substitute x = 0 and y = 15 and solve for a. then you can rewrite the equation of the parabola in standard form that's the 1st part hope it makes sense
once you have the equation in standard form you may need to use the general quadratic formula for the x intercepts
soo would it be y=x^2-8x+15
that's the correct equation... and the equation can be factored.
so what did you get for the x-intercepts
the equation is \[y=x^2 - 8x + 15~~~~then~~~~y=(x -5)(x-3)\]
so the intercepts are
where y = 0 so solve x - 5 = 0 and x - 3 = 0
ok.... so the x-intercepts these are points on the x-axis so y = 0 if you have a parabola to find the x-intercepts let y = 0 and solve for x so \[0 = x^2 - 8x + 15~~~or~~~0 =(x -5)(x -3)\] if either of the factors (x -5) or (x -3) are equal to zero the equation is equal to zero... so then you need to let the factors equal zero and solve for x so solve x - 5 = 0 and x - 3 = 0 what would you have...?
the other thing to think about is that the x-intercepts are the same distance from the x value in the vertex .
am go fail this test
so if you the solve the 2 equations what do you get x - 5 = 0 and x - 3 = 0
these values are the x-intercepts
correct so the intercepts are at (3,0) and (5, 0) the vertex has an x value of x = 4... the intercepts are the same distance from this value
thank u soo much
it was wrong
so did you write the intercepts as (3, 0) and (5, 0)..?