Given f(x) = x ^ 3 - x ^ 2 -4x +4
Find the zeros of f.
Write an equation of the line tangent to the graph of f at x = -1
The point (a,b) is on the graph of f and the line tangent to the graph at (a,b) passes through the point (0,-8) which is not of the graph of f. Find a and b
Please help. missing three classes in a row kills
this is my question linda goes to the grocery store every six days,jesus goes to the gorcery store every nine days. they both went to the grocery store today. in how many days will they both be at the grocery store again in the same day i cant solve it could you help me/give me the answer.please
your function factors by grouping
f(x) = x ^ 3 - x ^ 2 -4x +4
= x^2(x-1) - 4(x-1)
= (x-1)(x^2 - 1)
= (x-1)(x + 2)(x - 2)
so the function has zeros at x = 1,2, -2
f '(x) = 3x^2 - 2x - 4
f '(-1) = 3+2-4 = 1 , which is the slope
f(-1) = 3(-1) - (-1)^2 - 4(-1) + 4
= -3 - 1 + 4 + 4 = 4
so the point is (-1,4) and the slope is 1
can you take it from here?
Please do not post a new question as a reply to somebody else's question.
re yours ...
Linda goes on day 6, day 12, day 18, day 24 , ....
Jesus goes on day 9, day 18, day 27, ...
mmmmhhhh ?
To find the zeros of a function, you need to solve the equation f(x) = 0. In this case, the given function is f(x) = x^3 - x^2 - 4x + 4, so we need to solve the equation x^3 - x^2 - 4x + 4 = 0.
One way to find the zeros is by using factoring. By testing some integer values, we can determine that x = 1 is a solution of the equation. Using synthetic division or long division, we can divide the polynomial f(x) by (x-1), which gives us a quotient of x^2 - 4x + 4. This can be further factored as (x-2)(x-2). Therefore, the zeros of f(x) are x = 1 and x = 2 (a double root).
To write an equation of the tangent line to the graph of f at x = -1, we can use the concept of the derivative. The derivative of f(x) gives us the slope of the tangent line. So let's find the derivative of f(x) and substitute x = -1 into it.
f'(x) = 3x^2 - 2x - 4
Plugging in x = -1, we get:
f'(-1) = 3(-1)^2 - 2(-1) - 4 = 3 + 2 - 4 = 1
Therefore, the slope of the tangent line at x = -1 is 1. Now, we need to find the y-coordinate (b) of the point on the graph. We can substitute x = -1 into the original function f(x) to obtain the value of f(-1).
f(-1) = (-1)^3 - (-1)^2 - 4(-1) + 4 = -1 - 1 + 4 + 4 = 6
So we have the point (-1, 6) on the graph. Now we can use the point-slope form of a line to write the equation of the tangent line:
y - 6 = 1(x - (-1))
y - 6 = x + 1
y = x + 7
Therefore, the equation of the line tangent to the graph of f at x = -1 is y = x + 7.
Finally, to find the point (a, b) on the graph of f where the tangent line at that point passes through (0, -8), we need to solve the simultaneous equations formed by the equations of the tangent line and the given point.
From the equation of the tangent line, we know that y = x + 7. Substituting x = 0, we get y = 0 + 7 = 7. This implies that the point (0, 7) is on the tangent line.
To find the value of x where this tangent line intersects the function f, we can substitute y = 7 and solve the equation f(x) = 7:
x^3 - x^2 - 4x + 4 = 7
By rearranging the equation and solving for x, we find that x = 2 is the solution.
Therefore, the point (a, b) on the graph of f where the tangent line passes through (0, -8) would be (2, 7).
I hope this helps! Let me know if you have any further questions.