Problem 107
Question
What can you say about the inflection points of a quadratic curve \(y=a x^{2}+b x+c, a \neq 0 ?\) Give reasons for your answer.
Step-by-Step Solution
Verified Answer
A quadratic curve has no inflection points because its second derivative is constant and non-zero.
1Step 1: Understand the Definition of Inflection Points
An inflection point is where the curve changes concavity — from concave up (cup-shaped) to concave down (cap-shaped) or vice versa. To find an inflection point, we typically look for where the second derivative of a function equals zero and changes sign.
2Step 2: Find the Second Derivative of the Quadratic Function
Given the function \[y = ax^2 + bx + c,\]the first derivative is \[y' = 2ax + b.\]The second derivative then becomes \[y'' = 2a.\]
3Step 3: Analyze the Second Derivative
The second derivative, \(y'' = 2a\), is a constant. It does not depend on \(x\), meaning the concavity of the quadratic curve (determined by the sign of \(2a\)) does not change. Therefore, there is no point on the curve where the concavity changes.
4Step 4: Conclusion About Inflection Points
Since the second derivative is constant and non-zero, the quadratic function does not have any inflection points. A quadratic function is either concave up everywhere if \(a > 0\) or concave down everywhere if \(a < 0\).
Key Concepts
Quadratic CurveSecond DerivativeConcavity
Quadratic Curve
When we discuss quadratic curves, we refer to functions of the form \[ y = ax^2 + bx + c \] where \(a eq 0\). These curves are parabolic in shape. Because of the square term, the graph of a quadratic function forms a unique "U" shape, either opening upwards or downwards. This shape distinguishes quadratic curves from linear or cubic curves.
- If \(a > 0\), the parabola opens upwards (like a smile) - a shape called concave up.
- If \(a < 0\), the parabola opens downwards (like a frown) - a shape called concave down.
Second Derivative
To understand properties like concavity of a curve, we need to delve into derivatives. For a quadratic function \[ y = ax^2 + bx + c, \] we start by finding its derivative. The first derivative \[ y' = 2ax + b \] gives us the slope or the rate of change of the curve at any point.
Taking the derivative of \(y'\), we obtain the second derivative: \[ y'' = 2a. \] This calculation reveals that for a quadratic function, the second derivative is a constant. It does not depend on \(x\), unlike more complex functions.
Taking the derivative of \(y'\), we obtain the second derivative: \[ y'' = 2a. \] This calculation reveals that for a quadratic function, the second derivative is a constant. It does not depend on \(x\), unlike more complex functions.
- The value of \(2a\) indicates the consistent "direction" of the concavity.
- A constant second derivative means the curve does not have any flexibility or points of changing concavity.
Concavity
Concavity deals with the "bending" nature of the curve. We say a curve is concave up when it bends upwards like a cup opening to the sky. It's concave down when it arches downwards like an upside-down bowl. To assess where a curve changes from concave up to concave down, we rely on the second derivative.
Since for a quadratic function \[ y'' = 2a \] is non-zero and constant, the function maintains either concavity up or down across its entire domain. This behavior is predictable once we determine the sign of \(a\):
Since for a quadratic function \[ y'' = 2a \] is non-zero and constant, the function maintains either concavity up or down across its entire domain. This behavior is predictable once we determine the sign of \(a\):
- If \(a > 0\), it is concave up everywhere – always forming a smile.
- If \(a < 0\), it is concave down everywhere – always forming a frown.
Other exercises in this chapter
Problem 106
a. Find the coordinates of the vertex of the parabola$$y=a x^{2}+b x+c, a \neq 0$$. b. When is the parabola concave up? Concave down? Give reasons for your answ
View solution Problem 107
Use a CAS to solve the initial value problems. Plot the solution curves. $$y^{\prime}=\cos ^{2} x+\sin x, \quad y(\pi)=1$$
View solution Problem 108
Use a CAS to solve the initial value problems. Plot the solution curves.$$y^{\prime}=\frac{1}{x}+x, \quad y(1)=-1$$
View solution Problem 108
What can you say about the inflection points of a cubic curve \(y=a x^{3}+b x^{2}+c x+d, a \neq 0 ?\) Give reasons for your answer.
View solution