Tangent line through arbitrary point?

In summary, for the given function f(x)=x2+5x+5 and point A(4,5), the task is to find a formula for the slope of a line passing through point A and an arbitrary point on the function. This can be done by finding the slope of the line through A and any other point on the parabola, which can be expressed as f'(x)=2x+5. The slopes of the lines through A and points on the parabola are 1 and 25, but the question is unclear on what it is asking for in part two.
  • #1
PandaherO
10
0

Homework Statement



Consider the function f(x)=x2+5x+5 and the point A(4,5).

Find a formula for the slope of a line passing through point A, and an arbitrary point on the function. Your answer should be a formula in terms of x.
slope=?

Homework Equations



f′(x)=2x+5

The Attempt at a Solution



Using your answers from the two previous questions, find the slope(s) of the line(s) through point A that is (are) tangent to f(x)

the slopes are 1, and 25 however I don't understand what part two of the question is asking for...
(formula for slope passing through point A, and arbitrary point on the function)
 
Physics news on Phys.org
  • #2
PandaherO said:

Homework Statement



Consider the function f(x)=x2+5x+5 and the point A(4,5).

Find a formula for the slope of a line passing through point A, and an arbitrary point on the function. Your answer should be a formula in terms of x.
slope=?

Homework Equations



f′(x)=2x+5

The Attempt at a Solution



Using your answers from the two previous questions, find the slope(s) of the line(s) through point A that is (are) tangent to f(x)

the slopes are 1, and 25 however I don't understand what part two of the question is asking for...
(formula for slope passing through point A, and arbitrary point on the function)

First off, your thread title seems misleading to me, as this problem has nothing to do with tangent lines. All they are asking for is the slope of a line that goes through A(4, 5) and an arbitrary point on the graph of y = x2 + 4x + 5. Note that A is not on the graph of this parabola.

The question boils down to finding the slope of the line through two points: A(4, 5) and some point on the graph of the parabola. What are the coordinates of each point on the parabola?
 

Related to Tangent line through arbitrary point?

1. What is a tangent line through an arbitrary point?

A tangent line through an arbitrary point is a straight line that touches a curve at a specific point, but does not intersect the curve at any other points. It is also known as a tangent to a curve at a given point.

2. How is a tangent line through an arbitrary point different from a secant line?

A secant line is a straight line that intersects a curve at two or more points, while a tangent line only touches the curve at one point. Additionally, a secant line can be drawn between any two points on a curve, while a tangent line can only be drawn at a specific point.

3. What is the equation for a tangent line through an arbitrary point?

The equation for a tangent line through an arbitrary point (x0, y0) on a curve with equation y = f(x) is y - y0 = f'(x0)(x - x0), where f'(x) is the derivative of the curve at the point x0.

4. How do you find the slope of a tangent line through an arbitrary point?

The slope of a tangent line through an arbitrary point can be found by taking the derivative of the curve at that point. This will give you the value of the slope at that point, which can then be used in the point-slope form of the equation for a line to find the equation of the tangent line.

5. Can a tangent line through an arbitrary point be horizontal or vertical?

Yes, a tangent line through an arbitrary point can be horizontal or vertical. A horizontal tangent line has a slope of 0 and a vertical tangent line has an undefined slope. This occurs when the derivative of the curve at that point is equal to 0 or undefined, respectively.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
812
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
332
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
4K
  • Calculus and Beyond Homework Help
Replies
1
Views
910
  • Calculus and Beyond Homework Help
Replies
4
Views
320
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
13
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Back
Top