The domain is all real numbers but I need the range. HELP!!
f(x)= -x^2 +3x-5
2007-12-09 09:36:09 · 4 answers · asked by Sponge 3 in Science & Mathematics ➔ Mathematics
How do you figure out the range?
2007-12-09 09:39:55 · update #1
Find the y value of the vertex of this parabola. Since it points down, the range will be less than or equal to this y value.
Vertex is at (-b/2a, f(-b/2a)). -b/2a = -3/2*-1 = -3/-2 = 3/2
f(3/2) = -(3/2)^2 + 3(3/2) - 5 = -9/4 + 9/2 - 5 = 9/4 - 5 = -11/4.
ANSWER: Range: {y | y is less than or equal to -11/4}.
Hope that helps!
2007-12-09 09:45:22 · answer #1 · answered by Anonymous · 0⤊ 0⤋
f(x)= -x^2 +3x-5
fx + x2 – 3x + 5 = 0 - In order to solve this non-linear equation, we need to move all the terms to the left side. In the problem: -terms - x2, 3x, -5 will be moved to the left side.
Notice that a term changes sign when it "moves" from one side of the equation to the other.
x E 0 - This is the solution set for the given equation.
2007-12-09 09:56:32 · answer #2 · answered by Hillke 1 · 0⤊ 0⤋
5/9+4(4/9)^2
2015-03-13 12:09:25 · answer #3 · answered by Mariaelena 1 · 0⤊ 0⤋
f(x) = -(x-1.5)^2 -2.75 so the range is x<= -2,75
(completing the square)
2007-12-09 09:46:08 · answer #4 · answered by santmann2002 7 · 0⤊ 0⤋