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nadezda [96]
3 years ago
12

Consider the line: y= -7/2x-7 find the equation of the line that is perpendicular and parallel to this line that passes through

the point (-2,5)
Mathematics
1 answer:
saveliy_v [14]3 years ago
3 0

Answer:

=5

Step-by-step explanation:

(5)=-7/2(-2)-7

5=7-7

=5

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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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the parabola has maximum at 9, meaning is a vertical parabola and it opens downwards.

it has a symmetry at x = -5, namely its vertex's x-coordinate is -5.

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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