+0  
 
0
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6. Let $a$ and $b$ be real numbers, where $a < b$, and let $A = (a,a^2)$ and $B = (b,b^2)$. The line $\(\overline{AB}\)$ (meaning the unique line that contains the point $A$ and the point $B$) has $x$-intercept $(-3/2,0)$ and $y$-intercept $(0,3)$. Find $a$ and $b$. Express your answer as the ordered pair $(a,b)$.

 

7. Find all points $(x,y)$ that are 13 units away from the point $(2,7)$ and that lie on the line $x - 2y = 10$. Give your answer as a list of points separated by semicolons, with the points ordered such that their $x$-coordinates are in increasing order. (So "(1,-3); (2,3); (5,-7)" - without the quotes - is a valid answer format.)

 

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

 

9. Assume that $f(3) = 4$. Name a point that must be on the graph of \(y=\tfrac12f\left(\tfrac{x}{2}\right)\).

Lightning  Jul 24, 2018
 #1
avatar
-1

Please get rid of the dollar signs.

Guest Jul 24, 2018
 #2
avatar+92430 
+2

7. Find all points (x,y) that are 13 units away from the point (2,7) and that lie on the line x - 2y = 10. Give your answer as a list of points separated by semicolons, with the points ordered such that their $x$-coordinates are in increasing order. (So "(1,-3); (2,3); (5,-7)" - without the quotes - is a valid answer format.)

 

Rewrite the equation of the line as   x  = 2y + 10       (1)

We can envision  that (2,7)  is the center  of a circle with a radius of 13

The equation of this circle is  (x - 2)^2  + ( y - 7)^2  = 13^2        (2)

 

Sub  (1)  into (2) for x  and we have

 

( 2y + 10 - 2)^2  + ( y - 7)^2  = 13^2   simplify

 

( 2y + 8)^2  + ( y - 7)^2   = 169

 

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

 

5y^2 + 18y - 56  = 0        factor as

 

(5y + 28) ( y - 2)  =  0

 

Setting each factor  to 0 and solving for y produces  y = -28/5    and  y  = 2

 

Sub these back into (1)  to find the associated x values

 

x = 2(-28/5) + 10  =  -56/5 + 50/5 =  -6/5

x = 2(2) + 10   =  14

 

So...the points on the line  x - 2y = 10 that are 13  units away from (2,7)  are   

 

(-6/5, -28/5)  and  ( 14, 2)

 

Here's a graph : https://www.desmos.com/calculator/37bqbov5mv

 

 

cool cool cool

CPhill  Jul 24, 2018
 #3
avatar+92430 
+2

6. Let a and b be real numbers, where a < b, and let A = (a,a^2) and B = (b,b^2). The line AB (meaning the unique line that contains the point A and the point B) has x-intercept (-3/2,0) and y-intercept (0,3). Find a and b. Express your answer as the ordered pair (a,b).

 

We can  find the slope of the line  as  

 

[ 3 - 0 ] / [ 0 - -3/2] =  3 / (3/2)  =  2

And the equation of this line  is

y = 2x + 3     (1)

 

If   A  = (a, a^2)   and B  = (b,b^2)   then the function that would contain these points is  y  =  x^2   (2)

 

Setting  (1)  equal to (2), we have that

 

x^2  = 2x + 3

x^2 - 2x - 3  = 0     factor

(x - 3) ( x + 1)   = 0

 

Setting each factor to 0  and   solving for x produces  x = 3   and x  = -1

 

So....a  = -1  and  b  = 3  ⇒  (a,b)   = (-1,3)

 

Here's a graph : https://www.desmos.com/calculator/e5ys1u6aps

 

 

cool cool cool

CPhill  Jul 24, 2018
 #4
avatar+92430 
+2

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

 

The point  ( 3,4)   is on  the original graph

 

The graph  of  y = f(x) + 4     would shift this point  upward by  4  units

 

So....the point  ( 3, 4 + 4)  =  (3, 8)  is on  y  =f(x) + 4

 

 

cool cool cool

CPhill  Jul 24, 2018
edited by CPhill  Jul 24, 2018
 #5
avatar+92430 
+2

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

 

The  "x/2"  part  horizontally stretches the graph by a factor of 2   and the "1/2"  part out front vertically compresses the graph by a factor of 1/2

 

So....the point   ( 3, 4)  is on the original graph....and the point  ( 2*3, 1/2 * 4)  =  (6, 2)  is on  y =(1/2)f (x/2)

 

cool cool cool

CPhill  Jul 24, 2018
edited by CPhill  Jul 24, 2018

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