Range of Function

bkellymusic117

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Feb 13, 2014
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f(x)= -4x^2 + 12x - 5

I'm getting h= -4/-3 (4/3)

the correct answer is (-infinity,4]

When i plug in this i don't get 4 as the value of k.
 
f(x)= -4x^2 + 12x - 5

I'm getting h= -4/-3 (4/3)

the correct answer is (-infinity,4]

When i plug in this i don't get 4 as the value of k.

Again you posted incomplete problem statement.

Please post EXACT problem as it was given to you.
 
Again you posted incomplete problem statement.

Please post EXACT problem as it was given to you.

1. What is the range of the function f (x) = – 4x^2 + 12x – 5?


Huettenmueller, Rhonda (2014-01-01). College Algebra DeMYSTiFieD, 2nd Edition (Kindle Locations 5343-5345). McGraw-Hill Education. Kindle Edition.

That is the exact problem.
 
Last edited by a moderator:
f(x)= -4x^2 + 12x - 5

I'm getting h= -4/-3 (4/3)

the correct answer is (-infinity,4]

When i plug in this i don't get 4 as the value of k.

Hello bkelly:

The phrase in red is too vague to understand; it's better to show what you did. For example, you are apparently using some expression that contains symbols h and k, but we cannot see it.

Here is another approach.

If we think graphically about function f, then we realize from inspecting the quadratic polynomial that the graph is a parabola which opens downwards. This tells us that the y-coordinate of the vertex point is the largest number in the range of f. (The parabola continues downward forever, so there is no smallest number in the range.)

Therefore, you need only determine the y-coordinate of the vertex. There is a simple formula, to determine the x-coordinate:

x-coordinate of vertex = -b/(2a)

That's easy to remember because it is what's left of the Quadratic Formula when the Discriminant is zero.

Substituting the coefficients, we get -12/[(2)(-4)] = 3/2

Therefore, the largest value in the range of f is f(3/2).



If you would like to understand the other method (using something with h and k), please show us what you started with and explain your steps.

Also, check out the information in the summary page of our posting guidelines.

Cheers :cool:
 
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