Use the first derivative to find the equation of a quadratic function given tangent lines to the graph of this function.
In what follows, \( C \) represents the constant of integration where applicable.
Problem
a) Find the equation of a quadratic function whose graph is tangent at \( x = 1 \) to the line with slope 8, tangent at \( x = -2 \) to the line with slope \(-4\), and tangent to the horizontal line \( y = -8 \).
b) Find the equations of the tangent lines at \( x = 1 \) and \( x = -2 \).
c) Graph the quadratic function obtained and the 3 tangent lines in the same coordinate system and label the tangent lines and points of tangency.
Show Solution to Problem
a) The slope of the tangent to the graph of a function \( f \) is given by its first derivative. Let \( f \) be the quadratic function written in standard form:
\[ f(x) = ax^2 + bx + c \]The first derivative of \( f \) is:
\[ f'(x) = 2ax + b \]Since the slope of the tangent line equals the value of the derivative at the point of tangency, we can set up two equations from the tangent lines at \( x = 1 \) and \( x = -2 \):
\[ f'(1) = 2a(1) + b = 8 \] \[ f'(-2) = 2a(-2) + b = -4 \]Solving this system of linear equations:
\[ 2a + b = 8 \] \[ -4a + b = -4 \]Subtracting the second equation from the first gives \( 6a = 12 \implies a = 2 \). Substituting \( a = 2 \) into \( 2(2) + b = 8 \) yields \( b = 4 \).
The third tangent line \( y = -8 \) is horizontal, so its slope is \( 0 \). A horizontal line is tangent to a parabola at its vertex. The \( x \)-coordinate of the vertex \( h \) is found by setting \( f'(x) = 0 \):
\[ 2ah + b = 0 \implies h = -\frac{b}{2a} \]Substituting \( a = 2 \) and \( b = 4 \):
\[ h = -\frac{4}{2(2)} = -1 \]The graph of the quadratic function has its vertex at \( (-1, -8) \), meaning \( f(-1) = -8 \):
\[ f(-1) = a(-1)^2 + b(-1) + c = 2(1) + 4(-1) + c = -8 \] \[ 2 - 4 + c = -8 \implies -2 + c = -8 \implies c = -6 \]Thus, the quadratic function \( f \) is:
\[ f(x) = 2x^2 + 4x - 6 \]b) Now that we have the equation of the quadratic function, we find the \( y \)-coordinates of the points of tangency at \( x = 1 \) and \( x = -2 \):
- At \( x = 1 \): \( y = f(1) = 2(1)^2 + 4(1) - 6 = 0 \text{ (point } (1, 0) \text{)} \).
- At \( x = -2 \): \( y = f(-2) = 2(-2)^2 + 4(-2) - 6 = -6 \text{ (point } (-2, -6) \text{)} \).
Using point-slope form \( y - y_1 = m(x - x_1) \):
- Tangent at \( x = 1 \) (slope \( m = 8 \), passes through \( (1, 0) \)): \( y - 0 = 8(x - 1) \implies y = 8x - 8 \)
- Tangent at \( x = -2 \) (slope \( m = -4 \), passes through \( (-2, -6) \)): \( y - (-6) = -4(x - (-2)) \implies y + 6 = -4(x + 2) \implies y = -4x - 14 \)
c) Graphs of the quadratic function and all three tangent lines:
