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vekshin1
3 years ago
10

Find the vertex of the parabola y= x^2-2x-1 vertex (?,?)

Mathematics
2 answers:
nignag [31]3 years ago
7 0
y=a(x-h)^2+k \Rightarrow \hbox{vertex}=(h,k)\\\\
y=x^2-2x-1\\
y=x^2-2x+1-2\\
y=(x-1)^2-2 \Rightarrow \hbox{vertex}=(1,-2)
SpyIntel [72]3 years ago
5 0
y=ax^2+bx+c\\\\vertex:(x_v;\ y_v)\\\\x_v=\frac{-b}{2a}\ and\ y_v=f(x_v)


y=x^2-2x-1\\\\a=1;\ b=-2;\ c=-1\\\\x_v=\frac{-(-2)}{2\cdot1}=\frac{2}{2}=1\\\\y_v=1^2-2\cdot1-1=1-2-1=-2\\\\Answer:(1;-2)
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Answer:

<em>-1</em>

Step-by-step explanation:

1. A water wheel rung’s height as a function of time can be modeled by the equation:

h - 8 = -9 sin6t

(b) Determine the maximum height above the water for a rung.

Given the rung's height modeled by the equation;

h - 8 = -9 sin6t

h(t) = -9sin6t + 8

At maximum height, the velocity of the rung is zero;

dh/dt = 0

dh/dt = -54cos6t

-54cos6t = 0

cos6t = 0/-54

cos6t = 0

6t = cos^-1(0)

6t = 90

t = 90/6

t= 15

Substitute t = 15 into the expression to get the maximum height;

Recall:

h(t) = -9sin6t + 8

h(15) = -9sin6(15) + 8

h(15) = -9sin90 + 8

h(15) = -9(1)+8

h(15) = -9+8

<em>h(15) = -1</em>

<em>hence the maximum height above the water is -1</em>

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2 years ago
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3 years ago
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Irina-Kira [14]
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