Find the range of the function f(x)=

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Rewrite the numerator as

x ² + 3x + 5 = (x - 1/2)² + 4 (x - 1/2) + 27/4

Then

(x ² + 3x + 5) / (2x - 1) = 1/2 × (x ² + 3x + 5) / (x - 1/2)

… = 1/2 × ((x - 1/2)² + 4 (x - 1/2) + 27/4) / (x - 1/2)

… = 1/2 × ((x - 1/2) + 4 + 27 / (4 (x - 1/2)))

… = 1/2 x + 7/4 + 27 / (8 (x - 1/2))

which clearly has a non-removable singularity at x = 1/2, which is to say this function has a domain including including all real numbers except 1/2.

For every number other than x = 1/2, the function takes on every possible real numbers, since 1/2 x + 7/4 alone takes on all real numbers.

So:

domain = {x ∈ ℝ | x ≠ 1/2}

range = {x ∈ ℝ}

i think is g(1)=-1.

hope that i can

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?