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notsponge [240]
3 years ago
6

Graph 3x+y=-10 -x+3y=0

Mathematics
1 answer:
nikdorinn [45]3 years ago
3 0

The slope-intercept form of an equation of a line:

y=mx+b

m - slope

b - y-intercept

We have the equations in the standard form. Convert to the slope-intercept form:

3x+y=-10            <em>subtract 3x from both sides</em>

y=-3x-10

-x+3y=0             <em>add x to both sides</em>

3y=x             <em>divide both sides by 3</em>

y=\dfrac{1}{3}x

-----------------------------------------------------

We need only two points to plot a graph of each function.

y=-3x-10

for x = 0 and for x = -3:

y=-3(0)-10=0-10=-10\to(0,\ -10)\\\\y=-3(-3)-10=9-10=-1\to(-3,\ -1)

y=\dfrac{1}{3}x

for x = 0 and for x = -3:

y=\dfrac{1}{3}(0)=0\to(0,\ 0)\\\\y=\dfrac{1}{3}(-3)=-1\to(-3,\ -1)

Look at the picture.

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Suppose a parabola has an axis of symmetry at x=-5, a maximum height of 9, and passes through the point (-7,1). Write the equati
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check the picture below.

so then, we can pretty much tell its vertex is at (-5 , 9), and we also know it passes through (-7, 1)


\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \leftarrow \textit{using this one}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-5\\ k=9 \end{cases}\implies y=a[x-(-5)]^2+9\implies y=a(x+5)^2+9


\bf \textit{we also know that } \begin{cases} x=-7\\ y=1 \end{cases}\implies 1=a(-7+5)^2+9 \\\\\\ -8=a(-2)^2\implies -8=4a\implies \cfrac{-8}{4}=a\implies -2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-2(x+5)^2+9~\hfill

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