Answer the following function.
f(x)=2x^2-x-1
A. Is the point (-2,9) on the graph of f?
B. If x equals 2, what is fx? What point(S) are on the graph of f?
c. if f(x)= -1, what is x? what point(s) are on the graph of f?
d. what is the domain of f?
e. List the x-intercpts, if any, of the graph of f.
f. List the y-intercept if any, of the graph of f.
**Can someone please help me with this. I have no idea, i'm lost!!!
Do you realize that
f(x) = 2x^2 - x - 1 and y = 2x^2 - x - 1 say the same thing ?
A) sub in the values x = -2 and y - 9 and see if they satisfy the equation
LS = 2(-2)^2 - (-2) - 1
= 8 + 2 + 1 = 9
RS = 9 = LS
Yes , the point (-2,9) does lie on the graph.
B) replace x with 2 and evaluate
C) f(x) = -1 is the same as saying y = -1
so 2x^2 - x - 1 = -1
2x^2 - x = 0
x(2x-1) = 0
x = 0 or x = 1/2
the two corresponding points are (0,-1) and (1/2, -1)
D) the domain of a function are the values of x that you may use in the equation. Can't we use any x we feel like ? , so the domain is the set of real numbers
E) at the x-intercept, y = 0
2x^2 - x - 1 = 0
(2x + 1)(x - 1) = 0
x = -1/2 or x = 1 -----> the two x-intercepts
F) for the y-intercept , let x = 0
y = 0 - 0 - 1 = -1
No worries, I'm here to help! Let's tackle each question one by one, shall we?
A. Is the point (-2,9) on the graph of f?
To find out if (-2,9) is on the graph of f, we can substitute x = -2 into the equation:
f(-2) = 2(-2)^2 - (-2) - 1
f(-2) = 2(4) + 2 - 1
f(-2) = 8 + 2 - 1
f(-2) = 9
So, yes! The point (-2,9) is on the graph of f.
B. If x equals 2, what is f(x)? What point(s) are on the graph of f?
To find f(2), we substitute x = 2 into the equation:
f(2) = 2(2)^2 - 2 - 1
f(2) = 2(4) - 2 - 1
f(2) = 8 - 2 - 1
f(2) = 5
So, if x equals 2, f(x) equals 5. The point (2,5) is on the graph of f.
C. If f(x) = -1, what is x? What point(s) are on the graph of f?
To find x when f(x) = -1, we set f(x) = -1 and solve for x:
-1 = 2x^2 - x - 1
2x^2 - x - 1 + 1 = 0
2x^2 - x = 0
x(2x - 1) = 0
x = 0 or x = 1/2
So, when f(x) = -1, x can be 0 or 1/2. The points (0,-1) and (1/2, -1) are on the graph of f.
D. What is the domain of f?
The domain of f is the set of all possible x-values for which the function is defined. Since f is a polynomial function, it is defined for all real numbers. Therefore, the domain of f is (-∞,∞) or all real numbers.
E. List the x-intercepts, if any, of the graph of f.
To find the x-intercepts, we set f(x) = 0 and solve for x:
0 = 2x^2 - x - 1
We can use the quadratic formula to solve this equation, but alas, my quadratic formula is currently occupied with clown business. However, you can solve it using the quadratic formula yourself! Just remember that x-intercepts are points on the graph where f(x) intersects the x-axis.
F. List the y-intercept, if any, of the graph of f.
The y-intercept is the point where the graph intersects the y-axis. To find it, we set x = 0 in the equation for f(x):
f(0) = 2(0)^2 - 0 - 1
f(0) = 0 - 0 - 1
f(0) = -1
So, the y-intercept is (0,-1).
I hope that helps! If you have any more questions or need further assistance, feel free to ask.
A. To check if the point (-2,9) is on the graph of f, we substitute x = -2 into the function and see if we get y = 9 as the result.
Substituting x = -2 into f(x), we have:
f(-2) = 2(-2)^2 - (-2) - 1
= 2(4) + 2 - 1
= 8 + 2 - 1
= 9
Since f(-2) equals 9, the point (-2,9) is on the graph of f.
B. To find f(2), we substitute x = 2 into the function:
f(2) = 2(2)^2 - 2 - 1
= 2(4) - 2 - 1
= 8 - 2 - 1
= 5
So f(2) equals 5. The point (2,5) is on the graph of f.
C. To find x when f(x) equals -1, we set f(x) equal to -1 and solve for x:
-1 = 2x^2 - x - 1
Simplifying the equation, we have:
2x^2 - x = 0
Factoring out x, we get:
x(2x - 1) = 0
Setting each factor equal to zero, we have:
x = 0 or 2x - 1 = 0
If 2x - 1 = 0, then 2x = 1, and x = 1/2.
So x can be 0 or 1/2. The points (0, -1) and (1/2, -1) are on the graph of f.
D. The domain of a function is the set of all possible input values. In this case, since there are no restrictions or limitations given in the function, the domain of f is all real numbers.
E. To find the x-intercepts, we set y (or f(x)) equal to zero and solve for x:
0 = 2x^2 - x - 1
Using factoring, quadratic formula, or completing the square, we find that the equation does not factor easily. By using the quadratic formula, we have:
x = (-(-1) ± √((-1)^2 - 4(2)(-1))) / (2(2))
= (1 ± √(1 + 8)) / 4
= (1 ± √9) / 4
= (1 ± 3) / 4
So the x-intercepts are:
x = 4/4 = 1 or x = -2/4 = -1/2
The x-intercepts of the graph of f are 1 and -1/2.
F. To find the y-intercept, we substitute x = 0 into the function:
f(0) = 2(0)^2 - 0 - 1
= -1
So the y-intercept is -1. The graph of f intersects the y-axis at the point (0, -1).
Sure! I can help you with that. Let's go through each question step by step.
A. To check if the point (-2,9) is on the graph of f(x)=2x^2-x-1, we need to substitute x=-2 into the function and see if the resulting y-coordinate is 9. Let's calculate:
f(-2) = 2*(-2)^2 - (-2) - 1 = 9
Since the resulting y-coordinate is indeed 9, the point (-2, 9) is on the graph of f.
B. To find f(x) when x=2, we substitute x=2 into the function:
f(2) = 2*(2)^2 - (2) - 1 = 7
So f(2) = 7. This gives us one point on the graph of f, which is (2, 7).
C. To find the value of x when f(x)=-1, we need to solve the equation:
-1 = 2x^2 - x - 1
Simplifying this equation, we get:
2x^2 - x = 0
x(2x - 1) = 0
So either x = 0 or x = 1/2. These values of x give us two points on the graph of f, which are (0, -1) and (1/2, -1).
D. The domain of a function is the set of all possible input values (x-values). In this case, since f(x) is a polynomial function, the domain is all real numbers. Therefore, the domain of f is (-∞, ∞).
E. To find the x-intercepts of the graph of f, we need to find the values of x that make f(x) equal to 0. This means we need to solve the equation:
0 = 2x^2 - x - 1
To solve this quadratic equation, we can factor or use the quadratic formula. Factoring is not possible in this case, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = -1, and c = -1. Substituting these values into the formula, we get:
x = (-(-1) ± √((-1)^2 - 4(2)(-1))) / (2(2))
Simplifying further, we have:
x = (1 ± √(1 + 8)) / 4
x = (1 ± √9) / 4
x = (1 ± 3) / 4
So the x-intercepts of the graph of f are x = -1 and x = 1/2.
F. The y-intercept of a graph is the point where it intersects the y-axis, which is when x = 0. So to find the y-intercept of f, we substitute x = 0 into the function:
f(0) = 2*(0)^2 - (0) - 1 = -1
Therefore, the y-intercept of the graph of f is at (0, -1).
I hope this helps! Let me know if you have any further questions.