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
15 years ago
15 years ago
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?
15 years ago
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11 months ago
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.