1)Without graphing, describe the end behavior of the graph of f(x)= -5x^2-3x+1

as x starts from a negative value and increases the equation -5x^2-3x+1 curves up and peaks and comes down and passes the y-intercept at (0,1) to negative infinity.

I don't know what the answer is. I got it wrong, I would love some el helpo

2)If f(x)= -4x^2 and g(x)= 2/x. Find [g0f](x)

g(F(x))=g(-4x^2)= 2/(-4x^2)= -1/(2x)^2

-1/(2x)^2 = [g0f](x)

This is wrong too, maybe it's suppose to be 1/(-2x^2)? I don't really see what the difference is.

1. consider both ± ∞

as x is a huge negative number the x^2 would make it positive, but very large, multiplying by -5 makes it "hugely" negative. As x is large, both ±, the -3x+1 becomes insignificant.
so for large + x's the same thing happens, the function becomes "hugely" negative.
so the parabola drops down into the third quadrant and down into the fourth quadrant.
Is that what you want ?

for #2
you have 2/(-4x^2)= -1/(2x)^2
should have been -1/(2x^2)

you have the exponent of 2 outside the brackets, thus also squaring the 2 of the 2x, only the x is squared.

I guess that's what they want for #2 Thank you so much

1) To describe the end behavior of a graph without graphing it, you can analyze the leading term of the polynomial function. In the given function f(x) = -5x^2 - 3x + 1, the leading term is -5x^2. Since the coefficient of the leading term is negative, the graph will open downwards. As x approaches positive infinity, the graph will approach negative infinity. Therefore, the end behavior of the graph of f(x) is that as x increases without bound, the graph decreases without bound.

2) The notation [g0f](x) represents the composition of functions g(x) and f(x). To find [g0f](x), you need to substitute f(x) into g(x). Let's begin:

g(f(x)) = g(-4x^2)

Now, let's substitute this into the given function g(x) = 2/x:

g(-4x^2) = 2/(-4x^2)

To simplify, we can rewrite -4x^2 as -(2x)^2:

g(-4x^2) = 2/(-(2x)^2)

We can simplify the expression:

g(-4x^2) = 2/(-4x^2) = -1/(2x)^2

Therefore, the correct expression for [g0f](x) is -1/(2x)^2.

In your incorrect attempt, you wrote -1/(2x)^2 as 1/(-2x^2). The difference is the placement of the negative sign. Since there is a negative sign in front of the fraction, it should stay in the numerator, making it -1 instead of 1. Additionally, the term (2x)^2 should be preserved inside the parentheses instead of using -2x^2.