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Lines tangent to parabolas a. Find the derivative function \(f^{\prime}\) for the following functions \(f\) b. Find an equation of the line tangent to the graph of \(f\) at \((a, f(a))\) for the given value of \(a\) c. Graph \(f\) and the tangent line. $$f(x)=3 x^{2}+2 x-10 ; a=1$$

Short Answer

Expert verified
Question: Find the equation of the tangent line to the function \(f(x) = 3x^2 + 2x - 10\) at the point where \(a = 1\), and briefly describe the graph of the function and the tangent line. Answer: The equation of the tangent line to the function \(f(x) = 3x^2 + 2x - 10\) at the point where \(a = 1\) is \(y = 8x - 13\). The graph of the function is a parabola, and the tangent line touches the parabola at the point of tangency \((1, -5)\) without crossing it.

Step by step solution

01

Find the derivative function f'(x) for the given function f(x)

To find the derivative function, use the power rule and sum/difference rule for differentiation: $$f'(x) = \frac{d(3x^2 + 2x - 10)}{dx} = 6x + 2$$
02

Find the coordinates of the point of tangency (a, f(a))

We're given that \(a = 1\). Plug \(a\) into the original function \(f(x)\) to find the y-coordinate: $$f(a) = f(1) = 3(1)^2 + 2(1) - 10 = -5$$ So the point of tangency is \((1, -5)\).
03

Calculate the slope of the tangent line at the point of tangency

Plug \(a\) into the derivative function \(f'(x)\) to find the slope \(m\) of the tangent line at the point of tangency: $$m = f'(a) = f'(1) = 6(1) + 2 = 8$$
04

Find the equation of the tangent line

We now have the point of tangency \((1, -5)\) and the slope of the tangent line \(m = 8\). Use the point-slope form of a linear equation to find the equation of the tangent line: $$y - (-5) = 8(x - 1)$$ Simplify the equation: $$y + 5 = 8x - 8$$ $$y = 8x - 13$$
05

Graph the function and the tangent line

Graph the original function \(f(x) = 3x^2 + 2x - 10\) and the tangent line \(y = 8x - 13\). The point of tangency \((1, -5)\) should be on both graphs, and the tangent line should be touching the function at that point without crossing it.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

tangent line
When studying functions, a tangent line is an essential concept. It is a straight line that touches a curve at one single point without crossing it. This line represents the instantaneous rate of change or the slope of the function at that specific point. In other words, if you imagine zooming in close enough to the curve at the point of tangency, the curve would appear nearly straight, resembling that tangent line.

For solving problems involving tangent lines, it is key to use the derivative of the function. The derivative, which we'll explore later, gives us the slope of the tangent line at any point along the curve. Once we know the slope and the exact coordinates of the point of tangency, we can use the point-slope form to write the equation of the tangent line.

Using the point-slope equation:
  • We need the slope (from the derivative)
  • The coordinates of the point of tangency
This is summarized in the equation: \( y - y_1 = m(x - x_1) \), where \(m\) stands for the slope from the derivative, and \( (x_1, y_1) \) are the coordinates of the point.
power rule
The power rule is a simple yet powerful tool for finding derivatives of functions involving powers of the variable. It states that if you have a function ¥f(x) = x^n¥, then its derivative is given by multiplying the power by the coefficient and subtracting one from the power: \( f'(x) = n imes x^{n-1} \).

Applying the power rule makes it easy to differentiate polynomials and other expressions with powers quickly. In our example above, the function given was a polynomial: \( f(x) = 3x^2 + 2x - 10 \). By applying the power rule, we find the derivative for each term separately:
  • For \(3x^2\), derivative: \(3 imes 2x^{2-1} = 6x \)
  • For \(2x\), derivative: \(2 imes 1x^{1-1} = 2 \)
  • Constant \(-10\) becomes \(0\) upon differentiation
This results in the complete derivative function: \( f'(x) = 6x + 2 \).
The power rule is critical as it allows for swift differentiation of various types of mathematical functions, streamlining the process of finding tangent lines and analyzing function behavior.
parabola
A parabola is the graph of a quadratic function, characterized by its distinctive U-shape. The standard form of a quadratic function is given by: \( f(x) = ax^2 + bx + c \).

Parabolas can open upwards or downwards, depending on the coefficient of the \(x^2\) term:
  • If **a** is positive, the parabola opens upwards.
  • If **a** is negative, the parabola opens downwards.
The vertex of the parabola, which is its highest or lowest point, provides insight into the function's behavior. In the example function \( f(x) = 3x^2 + 2x - 10 \), the parabola opens upwards since the coefficient 3 is positive.

Analyzing parabolas is fundamental because they appear often in real-world contexts, from physics and engineering problems to economics and natural phenomena. By understanding a parabola's properties, we can predict the behavior of quadratic functions and solve problems involving them.
differentiation rules
Differentiation rules are guidelines that simplify finding the derivative of functions. By using these rules, we can calculate derivatives quickly and systematically. There are several primary differentiation rules that are helpful:
  • **Power Rule:** Already explained, it handles powers of \(x\).
  • **Sum/Difference Rule:** Allows differentiation of sums and differences of functions: \((f+g)' = f' + g'\) and \((f-g)' = f' - g' \).
  • **Constant Rule:** The derivative of a constant is zero.
  • **Constant Multiple Rule:** If a function is multiplied by a constant, its derivative is that constant times the derivative of the function: \((cf)' = c imes f' \).
Each of these differentiation rules plays a role in simplifying complex differentiation tasks. In our example, we used the power rule and the sum/difference rule together to find the derivative of the quadratic function. Understanding these rules allows you to tackle a variety of calculus problems efficiently, making the process of finding derivatives straightforward.

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Most popular questions from this chapter

a. Derive a formula for the second derivative, \(\frac{d^{2}}{d x^{2}}(f(g(x))).\) b. Use the formula in part (a) to calculate \(\frac{d^{2}}{d x^{2}}\left(\sin \left(3 x^{4}+5 x^{2}+2\right)\right)\).

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