Determine a quadratic function f(x)=ax^2 + bx + c if its graph passes through the point (2,19) and has a horizontal tangent at the point (-1, 8)
first of all you got 2 points, so
19 = 4a + 2b + c
8 = a - b + c
subtract them:
11 = 3a + 3b
f ' (x) = 2ax + b
at x = -1 this is zero, so
2a(-1) + b = 0
-2a + b = 0 or b = 2a
plug into 3a + 3b = 11
3a + 6a = 11
a = 11/9
b = 22/9
go into one of the first equations to solve for c
was expecting nicer numbers, check my arithmetic
if the tangent is horizontal, the slope is zero
so, you want 2ax+b = 0
At x = -1 that means
2a = b
Also, since (-1,8) is on the curve, that means that
a - 2a + c = 8
-a+c = 8
Since (2,19) is on the curve,
4a + 4a + c = 19
9a+c = 19
so, a = 11/9
b = 22/9
c = 83/9
f(x) = 11/9 x^2 + 22/9 x + 83/9
Or, eschewing calculus, we know the vertex is at (-1,8) so
y = a(x+1)^2 + 8
Plugging in (2,19) we get
9a+8 = 19
a = 11/9
y = 11/9 (x+1)^2 + 8
which is the same as we got above
To determine the quadratic function \(f(x) = ax^2 + bx + c\), we can make use of the given information.
1. The graph passes through the point (2, 19):
Substituting the x and y values into the equation, we get:
\(19 = a(2)^2 + b(2) + c\)
\(19 = 4a + 2b + c\) ---(Equation 1)
2. The graph has a horizontal tangent at the point (-1, 8):
A horizontal tangent means that the derivative of the function \(f'(x)\) is equal to zero at that point. Let's find the derivative of \(f(x)\):
\(f'(x) = 2ax + b\)
Substituting x = -1 and f'(x) = 0, we have:
\(0 = 2a(-1) + b\)
\(0 = -2a + b\) ---(Equation 2)
Now, we can solve the system of equations (Equation 1 and Equation 2) to find the values of a, b, and c.
From Equation 2, we can solve for b:
\(b = 2a\) ---(Equation 3)
Substituting Equation 3 into Equation 1, we get:
\(19 = 4a + 2(2a) + c\)
\(19 = 8a + c\) ---(Equation 4)
Now, we have two equations (Equation 3 and Equation 4) and two unknowns (a and c). We can solve this system of equations:
From Equation 3, substitute \(b = 2a\) into Equation 4:
\(19 = 8a + c\)
\(c = 19 - 8a\) ---(Equation 5)
Substituting Equation 5 into Equation 4, we have:
\(19 = 8a + (19 - 8a)\)
\(19 = 8a + 19 - 8a\)
\(19 = 19\)
The equation \(19 = 19\) is always true, which means that there are infinitely many solutions for \(a\) and \(c\). The value of \(b\) can be expressed in terms of \(a\) using Equation 3.
Therefore, the quadratic function \(f(x) = ax^2 + bx + c\) that passes through the point (2, 19) and has a horizontal tangent at the point (-1, 8) can be written as \(f(x) = ax^2 + 2ax + (19 - 8a)\), where \(a\) can be any real number.
To determine the quadratic function f(x) in the form f(x) = ax^2 + bx + c, we can use the given information about the graph. Let's break down the process step by step.
Step 1: Determine the values of a, b, and c.
Let's start by considering the point (2, 19) that the graph passes through. We can substitute the x and y coordinates into the quadratic equation to get an equation we can work with.
19 = a(2)^2 + b(2) + c
Simplifying this equation, we have:
19 = 4a + 2b + c
Step 2: Determine the slope at the point (-1, 8).
The graph has a horizontal tangent at the point (-1, 8). Since the tangent is horizontal, it implies that the derivative of the quadratic function f(x) with respect to x is zero at that point.
To find the derivative of f(x), we differentiate the quadratic function:
f'(x) = 2ax + b
Since the slope is zero at (x, y), we have:
0 = 2a(-1) + b
Simplifying this equation, we have:
0 = -2a + b
Step 3: Solve the system of equations.
We now have a system of two equations with three variables (a, b, and c). We can solve this system using the given equations:
19 = 4a + 2b + c
0 = -2a + b
We can begin by isolating b in the second equation:
b = 2a
Substituting this into the first equation, we get:
19 = 4a + 2(2a) + c
19 = 4a + 4a + c
19 = 8a + c
Now we have two equations with two variables:
8a + c = 19 (Equation 1)
b = 2a (Equation 2)
Step 4: Substitute values to determine the quadratic function.
Using Equation 2, we can express b in terms of a: b = 2a.
Substituting this back into Equation 1, we have:
8a + c = 19
Now, substitute the value of c from Equation 1 into the general form of the quadratic function:
f(x) = ax^2 + bx + c
f(x) = ax^2 + 2ax + (19 - 8a)
f(x) = ax^2 + 2ax + 19 - 8a
Therefore, the quadratic function that satisfies the given conditions is:
f(x) = ax^2 + 2ax + 19 - 8a, where a is a real number.