Problem 42
Question
In Problems 42-48, find the tangent line, in standard form, to \(y=\) \(f(x)\) at the indicated point. $$ y=3 x^{2}-4 x+7, \text { at } x=2 $$
Step-by-Step Solution
Verified Answer
The tangent line at \( x = 2 \) is \( 8x - y = 5 \).
1Step 1: Find the Derivative
To find the equation of the tangent line, we first need to find the derivative of the function, which represents the slope of the tangent line at any point. The given function is \( y = f(x) = 3x^2 - 4x + 7 \). By differentiating, we find the derivative: \( f'(x) = \frac{d}{dx}(3x^2 - 4x + 7) \). Using the power rule, we get \( f'(x) = 6x - 4 \).
2Step 2: Evaluate the Derivative at a Point
Next, we need to find the slope of the tangent line at the given point \( x = 2 \). Substitute \( x = 2 \) into the derivative: \( f'(2) = 6(2) - 4 = 12 - 4 = 8 \). So the slope of the tangent line is 8.
3Step 3: Calculate the Function Value at the Point
Substitute \( x = 2 \) into the original function to find the y-coordinate of the point where the tangent line touches the curve. \( y = f(2) = 3(2)^2 - 4(2) + 7 = 12 - 8 + 7 = 11 \). The point of tangency is \( (2, 11) \).
4Step 4: Write the Equation of the Tangent Line
Use the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point \( (2, 11) \). Substitute the values in: \( y - 11 = 8(x - 2) \).
5Step 5: Convert to Standard Form
Expand and simplify the equation obtained in Step 4 to convert it to standard form \( Ax + By = C \). Start by expanding: \( y - 11 = 8x - 16 \). Rearrange terms to get: \( 8x - y = 5 \). The standard form of the equation is \( 8x - y = 5 \).
Key Concepts
DerivativeSlope of the TangentPoint-Slope Form
Derivative
A derivative is a fundamental concept in calculus representing the rate of change of a function concerning its variable. In simple terms, the derivative tells us how the function's output value changes as its input value changes slightly. When we're working with a function like \( y = 3x^2 - 4x + 7 \), the derivative helps to find the slope of the tangent line to the curve at any given point.
To find the derivative \( f'(x) \), we apply differentiation rules such as the power rule. The power rule states that for a term \( ax^n \), the derivative is \( nax^{n-1} \). Applying this to the function, we get the derivative: \( f'(x) = 6x - 4 \). This derivative can now be used to determine the slope of the tangent line at any point along the curve.
If we set \( x = 2 \), substituting this into the derivative expression \( f'(x) \) gives us \( f'(2) = 8 \). This means the slope of the tangent line to the curve at \( x = 2 \) is 8. Understanding derivatives this way helps lay the groundwork for graph analysis, optimization, and motion problems.
To find the derivative \( f'(x) \), we apply differentiation rules such as the power rule. The power rule states that for a term \( ax^n \), the derivative is \( nax^{n-1} \). Applying this to the function, we get the derivative: \( f'(x) = 6x - 4 \). This derivative can now be used to determine the slope of the tangent line at any point along the curve.
If we set \( x = 2 \), substituting this into the derivative expression \( f'(x) \) gives us \( f'(2) = 8 \). This means the slope of the tangent line to the curve at \( x = 2 \) is 8. Understanding derivatives this way helps lay the groundwork for graph analysis, optimization, and motion problems.
Slope of the Tangent
The slope of the tangent line to a function at a particular point is the value of the derivative at that specific point. Imagine the tangent line as a straight line that just touches the curve at a given point without crossing it. This line reflects the momentary rate of change of the function at that precise location.
When you calculate the derivative of the function \( y = 3x^2 - 4x + 7 \) and then substitute \( x = 2 \), you find the slope of the tangent line, which is 8. This indicates that at \( x = 2 \), the function increases 8 units in the y-direction for every unit increase in the x-direction.
This slope is crucial because it gives insight into the behavior of the function at that point. If the slope were zero, it would mean the function has a horizontal tangent, indicating a potential maximum or minimum. A positive slope denotes increasing behavior, while a negative slope signifies decreasing behavior.
When you calculate the derivative of the function \( y = 3x^2 - 4x + 7 \) and then substitute \( x = 2 \), you find the slope of the tangent line, which is 8. This indicates that at \( x = 2 \), the function increases 8 units in the y-direction for every unit increase in the x-direction.
This slope is crucial because it gives insight into the behavior of the function at that point. If the slope were zero, it would mean the function has a horizontal tangent, indicating a potential maximum or minimum. A positive slope denotes increasing behavior, while a negative slope signifies decreasing behavior.
Point-Slope Form
The point-slope form of a linear equation is a tool you can use to write the equation of a line if you know a point on the line and its slope. The formula is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the given point.
For our exercise, with a slope \( m = 8 \) and a point \( (2, 11) \), we substitute these values into the point-slope form to get \( y - 11 = 8(x - 2) \). This form makes it easy to visualize the relationship between the point and the slope, as it directly shows how changes in \( x \) affect \( y \) given those specific conditions.
Once you have the equation in point-slope form, you might want to convert it to other forms, such as the standard form or slope-intercept form, depending on the problem's requirements. In this case, by simplifying and rearranging the terms, the equation becomes \( 8x - y = 5 \), which is in standard form. This is often more practical for solving and interpreting linear algebra problems.
For our exercise, with a slope \( m = 8 \) and a point \( (2, 11) \), we substitute these values into the point-slope form to get \( y - 11 = 8(x - 2) \). This form makes it easy to visualize the relationship between the point and the slope, as it directly shows how changes in \( x \) affect \( y \) given those specific conditions.
Once you have the equation in point-slope form, you might want to convert it to other forms, such as the standard form or slope-intercept form, depending on the problem's requirements. In this case, by simplifying and rearranging the terms, the equation becomes \( 8x - y = 5 \), which is in standard form. This is often more practical for solving and interpreting linear algebra problems.
Other exercises in this chapter
Problem 42
Assume that \(f(x)\) and \(g(x)\) are differentiable at \(x\). Find an expression for the derivative of \(y .\) $$ y=[f(x)-3] g(x) $$
View solution Problem 42
Differentiate the functions with respect to the independent variable. $$ g(s)=\exp \left[s^{2}+\ln s\right] $$
View solution Problem 43
Assume that the measurement of \(x\) is accurate within \(2 \% .\) In each case, determine the error \(\Delta f\) in the calculation of \(f\) and find the perce
View solution Problem 43
Find \(\frac{d y}{d x}\) by applying the chain rule repeatedly. \(y=\left(1+\left(3 x^{2}-1\right)^{3}\right)^{2}\)
View solution