find the equation of a quadratic function whose graph is tangent at x=1 to the line whose slope8, tangent at x=-2 to solve the line with slope-4 and tangent to the line y=-8
Start with defining the function:
f(x)=ax²+bx+c
Find its derivative:
f'(x)=2ax+b
1. f'(1)=8 =>
2a(1)+b = 8 ....(1)
2. f'(-2)=-4 =>
2a(-2)+b = -4 ...(2)
Solve system of equations (1) and (2) for a and b. (a=2,b=4)
The parabola is tangent to the line y=-8 => y=-8 is the minimum.
Now find the minimum of f(x) by setting f'(x)=0 and solve for x.
f'(x)=2ax+b=0
2(2)x+(4)=0
x=-1
So
f(-1)=-8
2(-1)²+4(-1)+c = -8
so c=-6
=>
f(x) = 2x²+4x-6
Do some checking to make sure the function satisfies all the conditions (and in case I make an arithmetic error):
given conditions to be checked:
f(1)=8
f(-2)=-4
minimum f(x)=-8
find the slope of the tangent line to the graph of e^xy=x at (1,0)
write an equation of the tangent line
To find the equation of a quadratic function that satisfies the given conditions, we need to use the information about the tangents and their slopes.
1. The graph of the quadratic function is tangent to a line at x = 1 with a slope of 8.
2. The graph is also tangent to a line at x = -2 with a slope of -4.
3. And, it is also tangent to the line y = -8.
Let's break it down step by step:
Step 1: Finding the equation based on tangents at x = 1 and x = -2.
The equation of a quadratic function in vertex form is: f(x) = a(x - h)^2 + k, where (h, k) is the vertex.
Since the graph is tangent to the line at x = 1 with a slope of 8, we know that the vertex is (1, f(1)). Let's denote f(1) as c.
We can rewrite the quadratic equation using the vertex form and replace f(1) with c:
f(x) = a(x - 1)^2 + c
Similarly, for the tangent at x = -2 with a slope of -4, the vertex is (-2, f(-2)), and we can write:
f(x) = a(x + 2)^2 + d
Step 2: Use the third given condition to eliminate the unknowns.
The graph is tangent to the line y = -8, so when x = 1, the function's value should equal -8.
Substitute x = 1 and set f(x) equal to -8:
a(1 - 1)^2 + c = -8
0 + c = -8
c = -8
This implies that the vertex is (1, -8), and we can rewrite the equation as:
f(x) = a(x - 1)^2 - 8
Step 3: Use the fact that the graph is tangent at x = -2 with slope -4 to determine the value of a.
The derivative of the quadratic function is: f'(x) = 2a(x - 1).
We know that the slope of the tangent at x = -2 is -4. So, when x = -2, f'(x) should equal -4:
f'(-2) = 2a(-2 - 1) = -4
-6a = -4
a = -4 / -6
a = 2/3
Step 4: Combine all the information and find the final equation.
Now that we know a = 2/3, we can substitute it back into the equation:
f(x) = (2/3)(x - 1)^2 - 8
This is the equation of the quadratic function that satisfies the given conditions.
To find the equation of a quadratic function whose graph is tangent at a specific point, we need to determine the values of the quadratic equation's coefficients.
Let's break the problem into steps:
Step 1: Tangent at x = 1 to a line with slope 8
When a quadratic function is tangent to a line at a specific point, the derivative of the quadratic function at that point will be equal to the slope of the tangent line.
Let's say the quadratic function is of the form f(x) = ax^2 + bx + c, where a, b, and c are coefficients to be found.
Since the function is tangent to a line at x = 1 with a slope of 8, the derivative of the quadratic function at x = 1 should be equal to 8.
Taking the derivative of f(x), we get:
f'(x) = 2ax + b
Plugging in the value x = 1 and setting it equal to the slope 8:
2a(1) + b = 8
2a + b = 8 --> Equation (1)
Step 2: Tangent at x = -2 to a line with slope -4
Similarly, we can set up another equation using the second point of tangency.
Using the derivative of f(x) again, we get:
f'(x) = 2ax + b
Plugging in x = -2 and setting it equal to the slope -4:
2a(-2) + b = -4
-4a + b = -4 --> Equation (2)
Step 3: Finding the equation of the tangent line
Given that the tangent line at x = 1 is y = 8x + k (since its slope is 8), we can find the value of k by substituting x = 1 and y = f(1) into the equation of the quadratic function:
f(1) = a(1)^2 + b(1) + c
f(1) = a + b + c
Since the line is tangent to the quadratic function at x = 1, the y-coordinate of the quadratic function at x = 1 should be equal to the corresponding y-coordinate of the tangent line (8). Therefore:
a + b + c = 8 --> Equation (3)
Step 4: Solving the system of equations
We now have a system of three equations with three unknowns (a, b, c):
Equation (1): 2a + b = 8
Equation (2): -4a + b = -4
Equation (3): a + b + c = 8
To solve this system of equations, we can use the method of substitution or elimination. Solving it using elimination:
Multiplying Equation (1) by 2, we get:
4a + 2b = 16 --> Equation (4)
Subtracting Equation (2) from Equation (4):
4a + 2b - (-4a + b) = 16 - (-4)
4a + 2b + 4a - b = 16 + 4
8a = 20
a = 20/8
a = 5/2
Substituting the value of a into Equation (1):
2(5/2) + b = 8
5 + b = 8
b = 8 - 5
b = 3
Substituting the values of a and b into Equation (3):
(5/2) + 3 + c = 8
(5/2) + 3 = 8 - c
(5/2) + 6/2 = 8 - c
11/2 = 8 - c
11/2 - 8 = -c
11/2 - 16/2 = -c
-5/2 = -c
c = 5/2
Step 5: Writing the equation of the quadratic function
Now that we have the values of a, b, and c, we can write the equation of the quadratic function:
f(x) = (5/2)x^2 + 3x + 5/2
Therefore, the equation of the quadratic function whose graph is tangent at x = 1 to the line with slope 8 and tangent at x = -2 to the line with slope -4 is:
f(x) = (5/2)x^2 + 3x + 5/2