needing help with tangents, line eqns, predictors

[rofl]

I cannot figure out my homework at all. My prof goes extremely fast and explains everything the hardest way possible. He doesnt even go over problems with the same format as the homework! Please help!!

1. Find all points of the curve y=f(x) at which the tangent line is horizontal

it gives you that y=10-x^2

2. Find all points of the curve y=f(x) at which the tangent line is horizontal

it gives you y=x^2-2x+1

3. One of the two lines that pass through the point (3,0) and are tangent to the parabola y=x^2 is the x-axis. Find an equation for the other line.

4. First apply the slope predictor formula in (10) for quadratic funtions to write the slope m(a) of the line tangent to y=f(x) at the point where x=a. Then write an equation of the line tangent to the graph of f at the points (2,f(2)).

it gives you f(x)=5

 

[wjm11]

1. Find all points of the curve y=f(x) at which the tangent line is horizontal

Find the derivative of the function, set it equal to zero, and solve for x.

 

[galactus]

I cannot figure out my homework at all. My prof goes extremely fast and explains everything the hardest way possible. He doesnt even go over problems with the same format as the homework! Please help!!

1. Find all points of the curve y=f(x) at which the tangent line is horizontal

it gives you that y=10-x^2

Differentiate, set to 0 and solve for x. Graph it will help. It's a parabola.
You don't even need calculus to find the max. Do you remember x=-b/2a
from algebra?. The x-coordinate of the vertex of the parabola.

2. Find all points of the curve y=f(x) at which the tangent line is horizontal

it gives you y=x^2-2x+1

Same as above

3. One of the two lines that pass through the point (3,0) and are tangent to the parabola y=x^2 is the x-axis. Find an equation for the other line.

This one is a little trickier than the other two. You don't know where it's tangent to the parabola.

You can use \(\displaystyle y-y_{1}=m(x-x_{1})\)

You are given y, y1, and x1.

The derivative(m=slope) of x^2 is 2x

\(\displaystyle \L\\\underbrace{x^{2}}_{\text{\uparrow\\y}}-\overbrace{0}^{\text{y1}}=\underbrace{2x}_{\text{m}}(x-\overbrace{3}^{\text{x1\\\downarrow}})\)


Solve for x. You can then use y=mx+b to find your line equation.

 

[rofl]

so then what would be the correct answer for the first one? i have sqrt10 but the book says its (0,10)

 

[stapel]

so then what would be the correct answer for the first one? i have sqrt10 but the book says its (0,10)

How did you get your answer? It would help if you showed your work and reasoning.

Please be complete. Thank you.

Eliz.

 

[confused_07]

For the first question on how to find a horizontal tangent line, how would you do that if you have not learned derivatives yet?