+1242 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 \(y=\tfrac12f\left(\tfrac{x}{2}\right)\).
+128407 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
+128407 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
+128407 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
+128407 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)
