answer the questions about the following function

f(x)= 10x^2/x^4+25

a. is the point (-sqrt 5,1) on the graph
b. if x=3, what is f(x)? what point is on the graph of f?
c. if f(x)=1, what is x? what points are on the graph?
d. what is the domain of f?
e. list the x-intersepts? if any on the graph of f.
f. list the y-inntercept, if any on the graph of f.

Please help me understand this, and show work please.

a.

When x = sqrt ( 5 )

f ( x ) = 10 x ^ 2 / ( x ^ 4 + 25 ) =

10 * sqrt ( 5 ) ^ 2 / ( sqrt ( 5 ) ^ 4 + 25 ) =

10 * 5 / ( 25 + 25 ) = 50 / 50 = 1

yes

b.

When x = 3

f ( x ) = 10 x ^ 2 / ( x ^ 4 + 25 ) =

10 x ^ 2 / ( x ^ 4 + 25 ) =

10 * 3 ^ 2 ( 3 ^ 4 + 25 ) =

10 * 9 / ( 81 + 25 ) =

90 / 106 = 2 * 45 / ( 2 * 53 ) = 45 / 53

c.

10 x ^ 2 / ( x ^ 4 + 25 ) = 1 Multiply both sides by ( x ^ 4 + 25 )

10 x ^ 2 = x ^ 4 + 25

Substitution:

x ^ 2 = z

x ^ 4 = z ^ 2

10 z = z ^ 2 + 25

Exact solutions are :

z = + OR - 5

x ^ 2 = z

x = + OR - sqrt ( 5 )

d.

x ^ 4 + 25 differently of = 0

x ^ 4 are olway positive and

x ^ 4 + 25 are olvay positive so Domain = all real numbers

e.

x-intersept = point where f ( x ) = 0

In this case 10 x ^ 2 = 0

Obviously x = 0 so

x -inntercept point ( 0 , 0 )

f.

y-inntercept = point where x = 0

In this case :

f ( x ) = 10 * 0 ^ 2 / ( 0 ^ 2 + 25 ) =

10 * 0 / ( 0 + 25 ) = 0 / 25 = 0

y-inntercept point also ( 0 , 0 )

P.S:

If don't know how to solve equation:

10 z = z ^ 2 + 25

In google type:
quadratic equation online

When you see list of results click on:

Free Online Quadratic Equation Solver:Solve by Quadratic Formula

When page be open in rectangle type:

10 z = z ^ 2 + 25

and click option solve it

You will see solution step-by-step

a. To determine if the point (-sqrt 5,1) is on the graph of f(x), substitute x = -sqrt 5 into the function and see if the equation holds true.

f(-sqrt 5) = 10(-sqrt 5)^2 / (-sqrt 5)^4 + 25
= 10(5) / (5^2) + 25
= 50/25 + 25
= 2 + 25
= 27

Since f(-sqrt 5) = 27 ≠ 1, the point (-sqrt 5,1) is not on the graph of f(x).

b. To find f(x) when x = 3, substitute x = 3 into the function.

f(3) = 10(3)^2 / (3^4) + 25
= 90 / 81 + 25
= 1.11 + 25
≈ 26.11

So, f(3) ≈ 26.11. The point (3, 26.11) is on the graph of f.

c. To find the value of x when f(x) = 1, set the function equal to 1 and solve for x.

1 = 10x^2 / (x^4 + 25)

To find x, cross multiply and rearrange the equation:

10x^2 = x^4 + 25
x^4 - 10x^2 + 25 = 0

This equation can be factored as:

(x^2 - 5)^2 = 0

Taking the square root of both sides:

x^2 - 5 = 0
x^2 = 5
x = ±sqrt(5)

So, x can be either +sqrt(5) or -sqrt(5). The points on the graph of f are (sqrt(5), 1) and (-sqrt(5), 1).

d. The domain of f is the set of all possible values of x, except those that make the denominator of the function equal to zero. In this case, the denominator is x^4 + 25. Since x^4 + 25 can never be zero for any real value of x, the domain of f is all real numbers.

e. To find the x-intercepts, we need to find the values of x where f(x) = 0. Set the function equal to zero and solve for x:

0 = 10x^2 / (x^4 + 25)

The numerator can never be zero, so the only way for the fraction to be zero is if the denominator is infinity. But since the denominator x^4 + 25 can never be zero, there are no x-intercepts on the graph of f.

f. To find the y-intercept, substitute x = 0 into the function:

f(0) = 10(0)^2 / (0^4 + 25)
= 0 / 25
= 0

So, the y-intercept is (0,0).

To answer these questions, we will need to analyze the given function, f(x) = (10x^2)/(x^4 + 25). Let's go through each question step by step:

a. To determine whether the point (-√5, 1) is on the graph of f, we need to substitute the x and y values of the point into the function and check if the equation holds true.

Plugging in x = -√5 into the function:
f(-√5) = (10(-√5)^2)/((-√5)^4 + 25) = 10(5)/(5^2 + 25) = 50/50 = 1.

Since f(-√5) = 1, the point (-√5, 1) is indeed on the graph of f.

b. To find f(3), we substitute x = 3 into the function:
f(3) = (10(3)^2)/(3^4 + 25) = (10*9)/(81 + 25) = 90/106 = 45/53.

Therefore, f(3) ≈ 0.8491. The point (3, 0.8491) is on the graph of f.

c. To solve f(x) = 1, we set the function equal to 1 and solve for x:
(10x^2)/(x^4 + 25) = 1.

To simplify the equation, we can multiply both sides by (x^4 + 25) to eliminate the denominator:
10x^2 = x^4 + 25.

Rearranging the equation and combining like terms:
x^4 - 10x^2 + 25 = 0.

This is a polynomial equation in terms of x^2. We can substitute a variable y = x^2, resulting in:
y^2 - 10y + 25 = 0.

Now, we can solve this quadratic equation:
(y - 5)(y - 5) = 0,
(y - 5)^2 = 0.

Therefore, y = 5. Since y = x^2, we have x^2 = 5, which means x = ±√5.

Thus, when f(x) = 1, x can take the values -√5 and √5. Therefore, the points (-√5, 1) and (√5, 1) are on the graph of f.

d. The domain of f is the set of all possible x-values for which the function is defined. In this case, the only restriction is that the denominator cannot be zero, as division by zero is undefined. So, the domain of f is all real numbers except the values that make the denominator (x^4 + 25) zero. Since x^4 + 25 = 0 has no real solutions, the domain of f is all real numbers.

e. To find the x-intercepts of the graph of f, we need to determine the values of x for which f(x) = 0. In other words, we solve the equation:
(10x^2)/(x^4 + 25) = 0.

The numerator being zero results in the function f(x) being zero. So, we can solve 10x^2 = 0. This equation has only one solution: x = 0.

Hence, the x-intercept of the graph of f is (0, 0).

f. The y-intercept is the point at which the graph of f intersects the y-axis. To find it, we substitute x = 0 into the function:
f(0) = (10(0)^2)/(0^4 + 25) = 0/25 = 0.

Thus, the y-intercept is (0, 0).

In summary:
a. The point (-√5, 1) is on the graph of f.
b. f(3) ≈ 0.8491, and the point (3, 0.8491) is on the graph of f.
c. When f(x) = 1, the values of x are -√5 and √5, resulting in the points (-√5, 1) and (√5, 1) on the graph of f.
d. The domain of f is all real numbers.
e. The x-intercept of the graph of f is (0, 0).
f. The y-intercept of the graph of f is (0, 0).