1. Define y-intercept of a line?

Mathematics

1 answer:

BartSMP [9]2 years ago

4 0

1. The y-intercept of a line is on a graph where a line crosses through the y axis.

2. b is the y- intercept and m is the slope. For example something using y=mx+b could be written like y=3x+2 if the y-intercept is 2 and the slope is 3.

3. The slope is what describes the direction and steepness of a line on a graph. Rise and run mean that when you find a slope you must first find the rise over run. You must go from one dot on a graph to the next by first going up or down to where that you are on the same number on the y axis that the next point is, next go on the x axis and go left or right until you have reached the next point. For example if you went up 3 and over 2, the rise would be 3 and the run would be 2. The slope in that situation would be 3/2.

Hope this helps you.

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Use the slope-intercept form of the linear equation to write the equation of the line with the given slope and y-intercept. Slop

motikmotik

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3 years ago

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Which expression is equivalent to16x2+81?

RideAnS [48]

16*X² +81

The answer is 3

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= 16x² +81

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3 years ago

Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7).

creativ13 [48]

So hmm check the picture below

we know the vertex is at the origin, and the focus point is below it, that means two things, the parabola is vertical and it's opening downwards

notice the distance "p", from the vertex to the focus point, is just 7 units, however, since the parabola is opening downwards, the "p" value will be negative, so p = -7

$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{4{{ p}}(y-{{ k}})=(x-{{ h}})^2 }\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
4{{ p}}(y-{{ k}})=(x-{{ h}})^2\quad 
\begin{cases}
h=0\\
k=0\\
p=-7
\end{cases}\implies 4(-7)(y-0)=(x-0)^2
\\\\\\
-28y=x^2\implies y=-\cfrac{1}{28}x^2$