Originally Posted by
BritneyPO I'm really desperate!
Graph this parabola
x=y^2 and y=-(x-2)^2
Write logbase2 64=6 as an exponential equation.
$\displaystyle x=y^2$
Set up a table of values for x and solve for y. Here are a few. Plot the points
Code:
x | y
-----
0 | 0
4 | +/-2
9 | +/-3
16| +/-4
The domain is all positive real numbers and zero. The parabola opens to the right with its vertex at (0, 0) and the x-axis is its axis of symmetry.
$\displaystyle y=-(x-2)^2$
Do the same thing here. Set up your table and plot some points
Code:
x | y
------
0 | -4
1 | -1
-1 | -9
2 | 0
-2 |-16
3 | -1
-3 |-25
This parabola opens downward because a<0. The vertex is at (2, 0) and the y-intercept is -4
$\displaystyle \log_264=6$ is the same as $\displaystyle 2^6=64$