Find the values of a,b, and c if the parabola y=a(x^2)+bx+c is tangent to the line y=-2x+3 at (2,-1) and has a critical point when x=3
y' = 2ax + b
we want y'(3) = 0, so
6a+b = 0
the line -2x+3 has slope -2, so
we want y'(2) = -2, so
4a+b = -2
and we have (a,b) = (1,-6)
y = x^2 - 6x + c
We know that y(2) = -1, so
-1 = 4 - 12 + c
so, c = 7
y = x^2 - 6x + 7
To find the values of a, b, and c, we can use the following information:
1. The parabola is tangent to the line y=-2x+3 at the point (2,-1).
2. The parabola has a critical point when x=3.
Let's solve this step-by-step:
Step 1: Finding the derivative of the parabola
To find the critical point, we need to find the derivative of the parabola function.
Differentiating y = a(x^2) + bx + c with respect to x, we get:
dy/dx = 2ax + b (1)
Step 2: Find the slope of the tangent line at (2, -1)
Since the parabola is tangent to the line y = -2x + 3 at the point (2, -1), the slope of the tangent line at that point should be the same as the slope of the line.
The slope of the tangent line at (2, -1) is -2.
Step 3: Set up the equations
Setting the derivative of the parabola equal to the slope of the tangent line, we have:
2ax + b = -2 (2)
Since the parabola has a critical point when x = 3, we know that the derivative should be zero at x = 3. Therefore, we have:
2a(3) + b = 0 (3)
Step 4: Solve the equations
We now have a system of equations (2) and (3) with two variables (a and b). We can solve these equations to find the values of a and b.
From equation (3), we can solve for a in terms of b:
a = -b/6
Substituting this value of a into equation (2), we have:
2(-b/6)(2) + b = -2
-b/3 + b = -2
-b + 3b = -6
2b = -6
b = -3
Substituting the value of b into equation (3), we can solve for a:
2a(3) + (-3) = 0
6a - 3 = 0
6a = 3
a = 3/6 = 1/2
Step 5: Finding the value of c
To find the value of c, we can substitute the values of a and b into the original equation of the parabola and solve for c.
Using the point (2, -1), we have:
-1 = a(2^2) + b(2) + c
-1 = (1/2)(4) + (-3)(2) + c
-1 = 2 - 6 + c
-1 = -4 + c
c = -1 + 4
c = 3
Therefore, the values of a, b, and c are:
a = 1/2,
b = -3,
c = 3
To find the values of a, b, and c, we need to use the information given about the points of tangency and critical point.
1. Tangency at point (2, -1):
When the parabola is tangent to the line y = -2x + 3 at the point (2, -1), two conditions must be met. First, the slopes of the parabola and the line should be equal at the tangent point. Second, the y-coordinates of the parabola and the line should be the same at the tangent point.
The given line has a slope of -2. To find the slope of the parabola at x = 2, we differentiate the equation y = a(x^2) + bx + c with respect to x.
Differentiating y = a(x^2) + bx + c:
dy/dx = 2ax + b
Setting x = 2 and equating the derivative to -2 gives us:
2a(2) + b = -2 (Equation 1)
Substituting x = 2 into the equation of the parabola and setting it equal to the equation of the line at (2, -1) gives us:
a(2^2) + b(2) + c = -2(2) + 3
4a + 2b + c = -1 (Equation 2)
2. Critical point at x = 3:
At the critical point of a parabola, the derivative is equal to zero. Setting x = 3 in the derivative function 2ax + b gives us:
2a(3) + b = 0 (Equation 3)
Now, we have a system of equations (Equations 1, 2, and 3) that we can solve simultaneously to find the values of a, b, and c.
Solving Equation 1:
2a(2) + b = -2
4a + b = -2 (Equation 4)
Solving Equation 2:
4a + 2b + c = -1 (Equation 5)
Solving Equation 3:
2a(3) + b = 0
6a + b = 0 (Equation 6)
Now, we have a system of three equations (Equations 4, 5, and 6) with three variables (a, b, c). We can solve it using any preferred method of linear algebra, such as substitution, elimination, or matrix methods.
By solving the system of equations, we can obtain the values of a, b, and c that satisfy the given conditions of the parabola.