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# Graphing Functions Help

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I'm struggling with these problems, and help would be greatly appreciated. Thank you!

1. What are the coordinates of the points where the graphs of f(x)=x^3-x^2+x+1 and g(x)=x^3+x^2+x-1 intersect?

2. Let a and b be real numbers, where a

3. Find all points (x,y) that are 13 units away from the point (2,7) and that lie on the line x-2y=10.

4a. Assume that f(3)=4. Name a point that must be on the graph of y=f(x)+4.

4b. Assume that f(3)=4. Name a point that must be on the graph of y= 1/2f(x/2).

Jun 24, 2019

#1
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1. What are the coordinates of the points where the graphs of f(x)=x^3-x^2+x+1 and g(x)=x^3+x^2+x-1 intersect?

Set these equal

x^3 - x^2 + x + 1  =  x^3 + x^2 + x - 1        subtract  x^3, x from both sides

-x^2  + 1 = x^2 - 1       add  x^2 to both sides....subtract 1 from both sides

0  = 2x^2 - 2

2x^2 - 2  = 0

x^2 - 1  = 0

x^2  = 1         take both roots

x = 1      or  x = -1

When x = 1 , y  =(1)^3 - (1)^2 + 1 + 1  =  2

So (1, 2)  is one intersection point

And when x = - 1 , y =  (-1)^3 - (-1)^2  - 1 + 1  =  -2

So  (-1,-2)  is the other intersection point

See the graph here : https://www.desmos.com/calculator/nnvu6b9qhr

EDIT TO CORRECT AN ERROR   Jun 24, 2019
edited by CPhill  Jun 25, 2019
#2
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3. Find all points (x,y) that are 13 units away from the point (2,7) and that lie on the line x-2y=10.

We can construct  a circle  centered at (2,7)  with a radius of 13

The equation of this circle is

(x - 2)^2 + (y - 7)^2  = 169       (1)

Rearrange the equation of the line  as  x = 2y+10     (2)

Put (2) int (1)  for  x     and we have

(2y + 10 - 2)^2 + ( y - 7)^2  = 169

(2y + 8)^2 + (y - 7)^2 = 169   simplify

4y^2  + 32y + 64  + y^2 - 14y + 49  = 169

5y^2 + 18y + 113 = 169       subtract 169 from both sides

5y^2 + 18y - 56   = 0     factor this

(5y  + 28) ( y - 2)  = 0

Set each factor to 0  and solve for y

5y + 28  = 0                     y  - 2  =  0

5y  = -28                           y  = 2

y  = -28/5

Put both y values back into the equation of the line to find their associated x coordinates

x - 2(-28/5)  = 10

x + 56/5  =  50/5

x = 50/5 - 56/5

x = -6/5

So one point  is  ( -6/5, -28/5)

And

x - 2(2)  = 10

x - 4  = 10

x = 14

And the other point is  (14, 2)   Jun 24, 2019
#3
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4a. Assume that f(3)=4. Name a point that must be on the graph of y=f(x)+4.

The graph of  y = f(x) + 4    shifts  f(x) up by 4 units....so....

f(x) + 4    =   (3, 4 + 4)   =  ( 3 , 8)   Jun 24, 2019
#4
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Wow great job!   Nickolas  Jun 24, 2019
#5
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4b. Assume that f(3)=4. Name a point that must be on the graph of y= 1/2f(x/2).

The point (3, 4)  is on the originalgraph

The "x/2"  horizontally "stretches" the graph by a factor of 2

The (1/2)   vertically  "compresses" the graph by  a factor  of 1/2

So....a point on  y = (1/2)f(x/2)  must be     (2*3 , (1/2)*4)  =  (6, 2)   Jun 24, 2019
#6
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I'm sorry, all the answers are incorrect except for 4a and 4b.

Jun 24, 2019
#7
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Sorry....I mis-copied (1)....see my edit

However...(3)  is correct....see the graph, here : https://www.desmos.com/calculator/zvlbv8gf1t   CPhill  Jun 25, 2019
#8
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Indeed, 3 is correct, read it wrong

Jun 26, 2019