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2 years ago

the standard form of the equation of a parabola is y=x^2 -8x + 29. what is the vertex form of the equation?

2 answers:

5 0

Y - 29 = x² - 8x ←I moved the constant to the other side
y - 29 + (-4)² = x² - 8x + (-4)² ← I just completed the square (b/2)² added to each side

y - 13 = (x - 4)² ← factored perfect square trinomial

y = (x - 4)² + 13 ← move constant back with binomial

tatuchka [14]2 years ago

4 0

The vertex is (0,21)

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Dmitrij [34]

Y = 6x
4x + y = 7

(substitute y = 6x in the ‘y’ spot in 4x + y = 7)
4x + 6x = 7
10x = 7
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(substitute x = 7/10 in the ‘x’ spot in either of the two equations)
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Answer:
x = 7/10
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3 years ago

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balu736 [363]

Answer:

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8 0

3 years ago

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Find the equation of all tangent lines having slope of -1 that are tangent to the curve y=(9)/(x+1)

fredd [130]

Answer:

$f(x)=\frac9{x+1}\\ f'(x)=-\frac9{(x+1)^2}\\ f'(x)=-1\ \iff\ -\frac9{(x+1)^2}=-1\ \to \ \frac9{(x+1)^2}=1\ \to \ (x+1)^2=9\\ |x+1|=3\ \to \ x+1=3\ \vee\ x+1=-3\\ x_1=2\ \vee\ x_2=-4\\ f(x_1)=f(2)=\frac9{2+1}=3\\ f(x_2)=f(-4)=\frac9{-4+1}=-3$

First tangent line:

$y=f'(x_1)\cdot (x-x_1)+f(x_1)\ \to \ y=-1(x-2)+3\ \to \ y=-x+5$

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Notice: slope of -1 means that both $f'(x_1), \ f'(x_2)$ are equal to -1, so $f'(x_1)=-1 \ and \ f'(x_2)=-1$

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2 years ago

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