What is the vertex of this parabola?
y-5 = x^2?

 (0,5)

Work Shown:

y-5 = x^2

y = x^2 + 5 ... add 5 to both sides

y = 1x^2 + 5

y = 1(x-0)^2 + 5 ... x replaced with (x-0)

The equation is now in the form y = a(x-h)^2 + k

We have a = 1, h = 0, k = 5.

The vertex is (h,k) = (0,5)

The value of 'a' determines how vertically stretched or compressed the parabola is. Also, if a > 0, then the parabola opens upward; or if a < 0, then the parabola opens downward.

1.   an exact solution requires a radical solution - no decimal approximation.2. (a)4x^3 = 756x^3 = 756/4 = 189x =   cube root of 189 =   5.74   to nearest hundredth answer(b)   189 = 3*3*3*7   so we can write the cube root of 189 as 3∛7 answerchecking (a) 4*(5.74^3 =   756.48   which is very close(b)   4* ((3∛3)^3   = 4*189   = 756   - spot on! 3. i would use solution b because the original question asked for an exact solution.

Can you upload a picture of the graph