Problem 51

Question

Find a function $y=a x^{2}+b x+c$ whose graph has an $x$ -intercept of $1,$ a $y$ -intercept of $-2,$ and a tangent line with a slope of -1 at the $y$ -intercept.

Step-by-Step Solution

Verified

Answer

The function is $ y = 3x^{2} - x - 2 $.

1Step 1: Use the x-intercept to find a relationship

We know that the graph of the function intersects the x-axis at x = 1. Therefore, the function value at x = 1 is zero. This gives us the equation:\[ a(1)^2 + b(1) + c = 0 \]which simplifies to:\[ a + b + c = 0 \]

2Step 2: Use the y-intercept to find another equation

The y-intercept of the function is the point where x = 0, and it intersects the y-axis at y = -2. Therefore, the function value at x = 0 is -2, which gives us the equation:\[ c = -2 \]

3Step 3: Use the slope of the tangent at the y-intercept

The slope of the tangent to the curve at the y-intercept (x = 0) is given as -1. The slope of the tangent is the derivative of the function at that point. The derivative of the function is:\[ y' = 2ax + b \]At x = 0, the slope is -1. Therefore, we have:\[ 2a(0) + b = -1 \]which simplifies to:\[ b = -1 \]

4Step 4: Substitute known values to find 'a'

Substitute the known values of b = -1 and c = -2 into the equation from Step 1:\[ a + (-1) - 2 = 0 \]This simplifies to:\[ a - 3 = 0 \]Solving for 'a' gives:\[ a = 3 \]

5Step 5: Form the final function

Now that we have 'a' = 3, 'b' = -1, and 'c' = -2, we can substitute these back into the general form of the quadratic function to find our desired function:\[ y = 3x^{2} - x - 2 \]

Key Concepts

Understanding the X-InterceptExploring the Y-InterceptThe Concept of a Tangent LineUnderstanding Derivatives in Quadratic Functions

Understanding the X-Intercept

To understand the x-intercept of a quadratic function, we need to consider the point where the graph crosses the x-axis. At this point, the value of the function is zero, meaning the y-coordinate is 0. For the function given in the problem, $y = ax^2 + bx + c$, to find the x-intercept, we set the equation equal to zero and solve for $x$.
In our exercise, the x-intercept is at $x = 1$, which means plugging $x = 1$ into the function gives us zero. The equation becomes:

$a(1)^2 + b(1) + c = 0$
Simplifying, $a + b + c = 0$

This relationship helps in finding the coefficients $a$, $b$, and $c$ when combined with other information about the function.

Exploring the Y-Intercept

The y-intercept of a function is the point where the graph crosses the y-axis. At this point, the x-coordinate equals zero. In a quadratic function $y = ax^2 + bx + c$, the y-intercept is simply $c$ because the terms with $x$ become zero when $x = 0$.
In the provided exercise, the y-intercept is $-2$. Thus, when $x = 0$, the function simplifies to $c = -2$:

This provides us directly with our $c$ value.

Understanding the y-intercept can help determine where the curve starts on the vertical axis and assists in shaping the entire graph.

The Concept of a Tangent Line

A tangent line is a straight line that just "touches" a curve at a particular point. The important property of a tangent line is that it has the same slope as the curve at that exact point. This slope is given by the derivative of the function.
In our problem, the tangent line at the y-intercept has a slope of $-1$. Because the y-intercept occurs at $x = 0$, we use the derivative to find this slope:

The derivative of $y = ax^2 + bx + c$ is $y' = 2ax + b$.
At $x = 0$, $y' = b$, and we are given $b = -1$.

Knowing this helps clarify both the shape of the graph and provide another critical equation for solving coefficients.

Understanding Derivatives in Quadratic Functions

Derivatives play a crucial role in understanding the behavior of functions, particularly with curves. The derivative of a function gives the slope of the tangent line to the graph at any point. For quadratic functions like $y = ax^2 + bx + c$, the derivative is given by $y' = 2ax + b$.
This expression helps determine the rate of change of the function with respect to $x$. In our specific problem:

The derivative $y'$ at $x = 0$ provides the slope $b = -1$.
This information then helps align the graph with known points and behaviors.

The derivative's role extends beyond finding slopes—it is key in solving equations involving curves and surfaces.

Problem 51

Problem 52

Other exercises in this chapter

Problem 51

Find $d^{2} y / d x^{2}$ $$y=x \cos (5 x)-\sin ^{2} x$$

View solution

Problem 51

Suppose that a function $f$ is differentiable at $x_{0}$ and that $f^{\prime}\left(x_{0}\right)>0 .$ Prove that there exists an open interval containing \

View solution

Problem 52

Find $d^{2} y / d x^{2}$ $$y=\sin \left(3 x^{2}\right)$$

View solution

Problem 52

Suppose that a function $f$ is differentiable at $x_{0}$ and $\mathrm{de}$ fine $g(x)=f(m x+b),$ where $m$ and $b$ are constants. Prove that if \(x_

View solution