What is the vertex of...f(x)=x^2-3x+11/4?

Question

I need help solving this problem...thanks!

Answer 1

I think you'd prefer the more algebra 1 solution to it ;)

-b/(2a), b = -3 and a = 1
You get 3/2 for the x-value of the vertex. Since the vertex is part of the graph, you can plug the x-value in and get the appropriate y-value (1/2). Your vertex is then (3/2, 1/2)

Edit: I'm pretty sure the other guy thought you meant 1 and 1/4, not 11/4.

Answer 2

Put the equation of the parabola into vertex form. To do that complete the square.

y = x² - 3x + 11/4

y + 9/4 = (x² - 3x + 9/4) + 11/4

y + 9/4 - 11/4 = (x - 3/2)²

y - 1/2 = (x - 3/2)²

The vertex (h,k) = (3/2, 1/2).

Answer 3

f(x) = x^2 - 3x + 11/4
Substituting y for f(x),
y = x^2 - 3x + 9/4 - 9/4 + 11/4
y = (x - 3/2)^2 + 2/4,
so the vertex is at (3/2,1/2)

Answer 4

Just derive your function: f'(x)=2x-3 and then find the root of this latter: x=3/2 and finally plug that value in f: f(3/2)=3/4. Thus your vertex is (3/2; 3/4).