Problem 30
Question
Find the equation of the tangent line to the curve \(y=x^{2}-\) \(3 x+1\) at the point \((2,-1)\)
Step-by-Step Solution
Verified Answer
y = x - 3
1Step 1: Identify the biological concept
Determine which biology area this addresses.
2Step 2: Recall relevant principles
Review key concepts.
3Step 3: Apply knowledge
Use biological reasoning.
4Step 4: State the answer
The answer is: y = x - 3
Key Concepts
DifferentiationDerivativeTangent Line
Differentiation
Differentiation might sound a bit complex, but it's really about calculating rates of change. Think of how a car's position changes over time. In mathematics, differentiation lets us understand how a function changes with respect to a variable. For the equation of our curve, which is \( y = x^2 - 3x + 1 \), we're interested in finding out how \( y \) changes as \( x \) changes.
To achieve this, we apply the differentiation process to break down the overall change into tiny parts. Each small change in \( x \) (called the "infinitesimal change") affects the value of \( y \). By gathering how these tiny changes accumulate, differentiation helps us draw the steepness, or the slope, of the curve at any point.
This slope guides us in finding the exact direction the curve is heading at that point.
To achieve this, we apply the differentiation process to break down the overall change into tiny parts. Each small change in \( x \) (called the "infinitesimal change") affects the value of \( y \). By gathering how these tiny changes accumulate, differentiation helps us draw the steepness, or the slope, of the curve at any point.
This slope guides us in finding the exact direction the curve is heading at that point.
Derivative
The derivative is a crucial concept in calculus, providing essential insights into the behavior of functions. For our given function, \( y = x^2 - 3x + 1 \), the derivative is found as \( y' = 2x - 3 \). Here's how it's done:
- The power rule: For \( x^2 \), the derivative is \( 2x \).
- The constant multiplier rule: For \(-3x\), the derivative is \(-3\).
- Constants themselves, like \( +1 \), disappear when taking the derivative.
This derivative, \( y' = 2x - 3 \), tells us the slope or steepness of the tangent to the curve at any value of \( x \). It’s like knowing the speed of a car at each moment of its journey rather than just the average speed. This information will be key when we find the tangent line, especially at the specified point (2, -1).
- The power rule: For \( x^2 \), the derivative is \( 2x \).
- The constant multiplier rule: For \(-3x\), the derivative is \(-3\).
- Constants themselves, like \( +1 \), disappear when taking the derivative.
This derivative, \( y' = 2x - 3 \), tells us the slope or steepness of the tangent to the curve at any value of \( x \). It’s like knowing the speed of a car at each moment of its journey rather than just the average speed. This information will be key when we find the tangent line, especially at the specified point (2, -1).
Tangent Line
The tangent line to a curve at a specific point is a straight line that just touches the curve at that point without crossing it. It represents the instantaneous direction of the curve. For our curve \( y = x^2 - 3x + 1 \), finding the tangent at the point \((2,-1)\) requires two things: a point and the slope.
We've already identified the point. Next, we use the derivative \( y' = 2x - 3 \) to find the slope of this tangent at \( x = 2 \). Plugging in, we find the slope \( m = 2(2) - 3 = 1 \).
Once we have the slope \( m \) and the point \((x_1, y_1) = (2, -1)\), we can use the point-slope form of a line:
- Formula: \( y - y_1 = m(x-x_1) \)
- Plugging in our values: \( y + 1 = 1(x-2) \)
- This simplifies to the equation of the tangent line: \( y = x - 3 \).
This line intersects the curve at (2, -1) and follows its direction exactly at that point.
We've already identified the point. Next, we use the derivative \( y' = 2x - 3 \) to find the slope of this tangent at \( x = 2 \). Plugging in, we find the slope \( m = 2(2) - 3 = 1 \).
Once we have the slope \( m \) and the point \((x_1, y_1) = (2, -1)\), we can use the point-slope form of a line:
- Formula: \( y - y_1 = m(x-x_1) \)
- Plugging in our values: \( y + 1 = 1(x-2) \)
- This simplifies to the equation of the tangent line: \( y = x - 3 \).
This line intersects the curve at (2, -1) and follows its direction exactly at that point.
Other exercises in this chapter
Problem 29
Differentiate $$ f(x)=(a x+1)^{3} $$ with respect to \(x\). Assume that \(a\) is a positive constant.
View solution Problem 30
Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=\left(1+\frac{1}{x}\right)^{1 / 4}\) at \(a=1\)
View solution Problem 30
Find the derivative with respect to the independent variable. $$ g(s)=\left(\cos ^{2} s-3 s^{2}\right)^{2} $$
View solution Problem 30
Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln \left(1-x^{3}\right) $$
View solution