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!!!
A. is f(-2) = 9? if it is, then the point (-2,9) is on the graph. Otherwise, it's not on the graph
B.Plug 2 into f(x)
f(2) = 2 * (2^2) - 2 - 1
The points on this graph are (2, f(2))
C. Solve
-1 = 2*x^2-x-1
You should find two solutions for x.
D. The domain of f is the range of values for x for which f(x) is real. In this case, the domain is -infinity to infinity
E. Find the solutions when x = 0:
2*0^2 - 0 - 1 = f(x=0)
F. Find the solutions when f(x) = 0
0 = 2*x^2 -x - 1
Solve for the two values of x that make this equation true
Sure! I'm here to help you. Let's go through each question step by step.
A. To check if the point (-2, 9) is on the graph of f, we need to substitute x = -2 into the function and see if it satisfies the equation f(x) = 2x^2 - x - 1.
Substituting x = -2 into the function, we get:
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 indeed 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, when x = 2, f(x) is equal to 5.
To find the point(s) on the graph of f, we need to substitute different values of x into the function and get the corresponding y-values. For example, if we substitute x = 0, we can find the y-intercept.
C. To find the value(s) of x when f(x) = -1, we need to solve the equation 2x^2 - x - 1 = -1.
2x^2 - x - 1 = -1
2x^2 - x = 0
x(2x - 1) = 0
This equation has two solutions:
x = 0 or 2x - 1 = 0, which means x = 1/2.
So, when f(x) = -1, we have two values of x, 0 and 1/2. To find the corresponding points on the graph of f, substitute these values into the function.
D. The domain of f is the set of all possible values for x. In this case, there are no restrictions stated in the function, so the domain of f is all real numbers.
E. To find the x-intercepts of the graph of f, we set f(x) equal to zero and solve for x. In other words, we find the values of x that make f(x) = 0.
2x^2 - x - 1 = 0
To solve this quadratic equation, we can factor or use the quadratic formula. However, in this case, the equation is not easily factored. So, we will use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In our equation, a = 2, b = -1, and c = -1.
Using the quadratic formula:
x = (-(-1) ± √((-1)^2 - 4(2)(-1))) / (2(2))
= (1 ± √(1 + 8)) / 4
= (1 ± √9) / 4
= (1 ± 3) / 4
This gives two possible x-intercepts:
x = (1 + 3) / 4 = 4/4 = 1
x = (1 - 3) / 4 = -2/4 = -1/2
So, the x-intercepts of the graph of f are x = 1 and x = -1/2.
F. The y-intercept is the value of f(0). To find it, we substitute x = 0 into the function:
f(0) = 2(0)^2 - 0 - 1
= 0 - 0 - 1
= -1
So, the y-intercept of the graph of f is -1.