+98 The parabola y=ax^2+k has a vertex (0,-4) and passes through the point (3,7). Find the equation.
Find a formula for the quadratic function whose graph has its vertex at (1,4) and its y-intercept at y-6.
+128475 y = ax^2 + k
If the vertexx is (0, -4), then k = -4
And the point (3,7) is on the parbola implies that
7 = a (3)^2 - 4 add 4 to both sides
11 = a (9) divide both sides by 9
11/9 = a
So.....the equation is
y = (11/9)x^2 - 4
+934 This is a correction nobody will care about but this question is in Algebra 1 not precalc.
