The transverse axis of a hyperbola is the x-axis, and the equation of the asymptotes is y=± (4)/(3)x
1 answer:
Given that the transverse axis of a hyperbola is the x-axis, the re-written in the standard form,
a = 3
b = 1
c = 12.
<h3>How do you solve for the standard form of the above equation?</h3>It is to be noted that the standard form of a linear equation is given as:
ax + by = c
Rewritten in the normal format,
y=± (4)/(3)x ⇒ y-4 = -3 (x-3)
Simplifying -3(x-3) by applying distributive property and we have:
-3x -3 * -3
Hence
y-4 = -3x -3 * -3
⇒ y-4 = -3x + 9
Collecting like terms on one side of the equation we have
3x + y - 4 = 9
⇒ 3x + y = 9+4
⇒ 3x + y = 12 = Standard Form
Therefore, in standard from,
a = 3
b = 1
c = 12
See the relevant Graph attached.
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Answer:
y will be in every single quadrant
Step-by-step explanation:
So we have the equation first we will have to look at the equation. It says that y is less than or equal to
since y is less than
the only place the shaded area where y can be is under the line that is drawn be the equation. When the equation is graphed the y-intercept will be on positive 1 it since slope is rise over run it will look something like the file attached to this. so under the line you can see every single quadrant so that is why it would be that way
