Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. Find a second-degree polynomial \(f(x)=a x^{2}+b x+c\) such that its graph has a tangent line with slope 10 at the point (2,7) and an \(x\) intercept at (1,0)

Short answer

The answer is the quadratic function \(f(x)=a x^{2}+b x+c\) with the values of a, b, and c found by solving the simultaneous equations derived from the problem's given conditions. This needs to be confirmed by checking that the graph of the function obtained meets the problem's mentioned conditions.

Step-by-step solution

Finding the derivative

Start by differentiating the quadratic function \(f(x)=a x^{2}+b x+c\), which will yield \(f'(x)=2a x +b\). The derivative of a function at a certain point gives the slope of the tangent line at that point.

Use conditions to obtain equations

Substitute x-value of the given tangent point in the derivative, i.e \(f'(2)=10\). This will yield an equation in a and b. Also, use the given x-intercept (1,0) in the original quadratic function to derive another equation. This will give \(f(1)=0\). Therefore, two equations are obtained based on the conditions provided in the problem.

Solve the simultaneous equations

Solve the simultaneous equations to find the values of a, b and c. This will define the required second-degree polynomial.

Checking the solution

To check the solution, substitute the values of a, b and c in the quadratic formula and plot the function. The graph should meet the criteria explained in step 2.

Key concepts

Derivative of Quadratic Function

Understanding how to find the derivative of a quadratic function is fundamental in calculus. A quadratic function is generally given as \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. The derivative of the function, denoted as \(f'(x)\), represents the rate at which \(f(x)\) changes with respect to \(x\). To calculate the derivative, we apply the power rule to each term that involves \(x\).
For the quadratic function, the derivative is \(f'(x) = 2ax + b\). This formula is derived by taking the exponent, multiplying it by the coefficient, and then reducing the exponent by one. For instance, the derivative of \(ax^2\) is \(2ax^1\), which simplifies to \(2ax\). The derivative is essential in finding the slope of tangent lines to the graph of the function at any given point, thereby describing the function's concavity and points of inflection.

Practical Example:

In our exercise, we differentiate \(f(x) = ax^2 + bx + c\) to find the slope of the tangent line at a specific point, such as (2,7). This process will contribute to understanding how the function behaves near that particular point.

Tangent Line Slope

The slope of a tangent line to a curve at a given point is a measure of the steepness of the line that just touches the curve at that point. For any differentiable function, the slope of the tangent line at a point is exactly the value of the derivative of the function at that same point.
In the context of a quadratic function, where the general form is \(f(x) = ax^2 + bx + c\), the slope of the tangent line at a specific point \((x_1, y_1)\) is \(f'(x_1)\), which is \(2ax_1 + b\). Having the slope allows us to write the equation of the tangent line, given a point through which it passes.

Importance in Our Exercise:

For the polynomial provided in the exercise, we find that the slope of the tangent line at the point (2,7) is 10. By applying the derivative, we can set up an equation \(2a(2) + b = 10\) to connect the slope with the coefficients of the quadratic function.

X-Intercept Calculation

The x-intercepts of a function are the points where the graph of the function crosses the x-axis. Put simply, these are the points where the function \(f(x)\) equals zero. For quadratic functions, which are parabolic in shape, there can be zero, one, or two x-intercepts, depending on how the parabola is positioned in relation to the x-axis.
To find the x-intercepts algebraically, we set the quadratic function equal to zero, that is \(ax^2 + bx + c = 0\), and solve for \(x\). This often involves factoring, completing the square, or applying the quadratic formula.

Relation to the Given Exercise:

In our exercise, we are given one x-intercept at the point (1,0), meaning when \(x = 1\), \(f(x) = 0\). This gives us an additional equation \(a(1)^2 + b(1) + c = 0\), which helps in solving for the coefficients of the quadratic function along with the derivative information.

Solving Simultaneous Equations

Solving simultaneous equations is a method used to find the values of variables that satisfy multiple equations at the same time. These equations are connected through shared variables. The solutions are found by making one variable the subject in one of the equations and substituting it into another, by equating coefficients, or by using methods such as substitution or elimination.
Simultaneous equations are crucial when dealing with problems that have more than one unknown quantity, and they provide a way to derive these unknown values from the relationships between them.

Applying to Our Context:

For this calculus problem, we have two equations derived from the given conditions of the quadratic function. The first is related to the slope of the tangent line \(2a(2) + b = 10\), and the second is from the x-intercept \(a(1)^2 + b(1) + c = 0\). By solving these two equations together, we can find the values of \(a\), \(b\), and \(c\) that define the second-degree polynomial meeting the required criteria.