12 points please help

Mathematics

1 answer:

likoan [24]3 years ago

8 0

Answer:

Step-by-step explanation:

How to tell if a function passes the horizontal line test: Draw the graph of the function. Next, draw a horizontal line through the graph. If the line intersects the graph in more than one point, then the function fails the horizontal line test, which, in turn, means that the function does not have an inverse.

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Write the equation the describes the parabola that passes through the point (-8,-21) and has a vertex of (-4,-5), in vertex form

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Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8

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First, let $(x,y)$ be a point in our parabola. Since we know that the focus of our parabola is the point (0,8), we are going to use the distance formula to find the distance between the two points:
$\sqrt{(x-0)^{2}+(y-8)^{2}}$

Next, we are going to find the distance between the directrix and the point in our parabola. Remember that the distance between a point (x,y) of a parabola and its directrix, $y=c$ , is: $|y-c|$ . Since our directrix is y=-8, the distance to our point will be:
$|y-(-8)|$
$|y+8|$

Now, we are going to equate those two distances, and square them to get rid of the square root and the absolute value:
$\sqrt{(x-0)^{2}+(y-8)^{2}}=|y+8|$
$(\sqrt{(x-0)^{2}+(y-8)^{2}})^{2}=|y+8|^{2}$
$(x-0)^{2}+(y-8)^{2}=y+8$

Finally, we can expand and solve for $y$ :
$x^2+y^2-16y+64=y^2+16y+64$
$x^{2}-16y=16y$
$32y=x^{2}$
$y= \frac{1}{32} x^2$

We can conclude that t<span>he standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8 is </span> $y= \frac{1}{32} x^{2}$