Problem 65 - Precalculus with Limits

Question

In Exercises 65-68, find an equation of the tangent line to the parabola at the given point, and find the \(x\)-intercept of the line. \(x^2 = 2y, (4, 8)\)

Step-by-Step Solution

Answer

The equation of the tangent line to the parabola at the point (4, 8) is \(y = 4x - 8\), and the x-intercept of the line is \(x = 2\).

Step 1: Derive the equation

Before deriving, rewrite the equation in the form \(y = f(x)\). The original equation is \(x^2 = 2y\), rewrite to \(y = x^2/2\). Now, derive the equation with respect to \(x\). Using the power rule, the derivative of \(y\) with respect to \(x\) is \(y' = x\). This gives the slope of the tangent line.

Step 2: Form the equation of the tangent line

We can use the point-slope form of the linear equation, which is \(y - y_1 = m(x - x_1)\). Here, \((x_1, y_1)\) is the point of tangency and m is the slope of the tangent line. In this case, \(m = x_1 = 4\) and \((x_1, y_1) = (4, 8)\). So, substituting these into the equation, we get \(y - 8 = 4 (x - 4)\). Simplifying this equation gives \(y = 4x - 8\).

Step 3: Find the x-intercept of the line

Setting \(y\) to 0 in the equation of the line will give us the \(x\)-intercept. So, substituting 0 into \(y = 4x - 8\) and solving for \(x\), we get \(x = 2\). This is the \(x\)-intercept of the line.

Key Concepts

Derivative

In mathematics, the derivative of a function represents the rate of change of the function's value with respect to one of its variables. In simpler terms, it tells us how much the y-value of a function changes as the x-value changes. For the given function \( y = \frac{x^2}{2} \), the derivative is found using differentiation rules.
To find the derivative of this parabola, we used the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \). For \( y = \frac{x^2}{2} \), the derivative \( y' \) or \( \frac{dy}{dx} \) is simply \( x \).
This derivative, \( y' = x \), provides the slope of the tangent line at any given point along the curve of the equation.

Slope of the tangent line

The slope of the tangent line is a crucial concept in calculus and geometry. It denotes the steepness or incline of the tangent line at a particular point on the curve. The tangent line simply touches the curve at one point and mirrors the curve's slope precisely at that point.

For the function \( y = \frac{x^2}{2} \), we've calculated its derivative as \( y' = x \). This means that the slope of the tangent at any point \(x\) on the curve is \( m = x \). At the point \((4, 8)\), the derivative \( y' = 4 \) indicates that the slope of the tangent line here is also 4.
The concept of slope being exactly equal to the derivative at a point is fundamental when dealing with curves and their tangents.

Point-slope form

Point-slope form is a way of expressing the equation of a line. It's particularly useful when we know one point on the line and the line's slope. The general formula for point-slope form is: \[ y - y_1 = m(x - x_1) \]Here, \((x_1, y_1)\) is the given point, and \(m\) is the slope of the line.
In our scenario, the point of tangency is \((4, 8)\) and the slope \(m\) is 4. Plugging these values into the formula gives us: \[ y - 8 = 4(x - 4) \]After simplifying, this equation yields: \[ y = 4x - 8 \]
With this approach, constructing the equation of a tangent or any line becomes straightforward when one point and the slope are known.