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1 year ago

Find the coordinates of the vertex of the following parabola algebraically. Write your answer as and x,y point

1 answer:

8 0

The equation of the parabola in the vertex form is y = (x - 3 $)^{2}$ - 5 with ( 3, -5) is the vertex of the parabola and 1 is the multiplier

In the above question, A parabolic equation is given as follows:

Y = x^2 - 6x + 4

The equation of the parabola in the vertex form is :

y = a (x - h $)^{2}$ + k

Where a is a multiplier in the equation and (h,k) are the coordinates of the vertex

So, in order to obtain this form, we will use the method of completing square :

Y = x^2 - 6x + 4

y = $x^{2}$ - 6x + (9 -9) + 4

y = (x - 3 $)^{2}$ + ( -9 + 4)

y = (x - 3 $)^{2}$ - 5

where, ( 3, -5) is the vertex of the parabola and 1 is the multiplier

Hence, The equation of the parabola in the vertex form is y = (x - 3 $)^{2}$ - 5 with ( 3, -5) is the vertex of the parabola and 1 is the multiplier

To learn more about, parabola, here

brainly.com/question/21685473

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3 years ago

For 15 hours of work, a plumber charges $435. If he charges the same amount for each hour of work, which table shows the correct

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2 years ago

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Integrate sin^-1(x) dx<br><br> please explain how to do it aswell ...?

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If you're using the app, try seeing this answer through your browser: brainly.com/question/2264253

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx\qquad\quad\checkmark}$

Trigonometric substitution:

$\mathsf{\theta=sin^{-1}(x)\qquad\qquad\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}}$

then,

$\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}$

So the integral $\mathsf{(ii)}$ becomes

$\mathsf{=\displaystyle\int\! \theta\,cos\,\theta\,d\theta\qquad\quad(ii)}$

Integrate $\mathsf{(ii)}$ by parts:

$\begin{array}{lcl} \mathsf{u=\theta}&\quad\Rightarrow\quad&\mathsf{du=d\theta}\\\\ \mathsf{dv=cos\,\theta\,d\theta}&\quad\Leftarrow\quad&\mathsf{v=sin\,\theta} \end{array}\\\\\\\\ \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-\int\!sin\,\theta\,d\theta}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-(-cos\,\theta)+C}$

$\mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta+cos\,\theta+C}$

Substitute back for the variable x, and you get

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=sin^{-1}(x)\cdot x+\sqrt{1-x^2}+C}\\\\\\\\ \therefore~~\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=x\cdot\,sin^{-1}(x)+\sqrt{1-x^2}+C\qquad\quad\checkmark}$

I hope this helps. =)

Tags: <em>integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus</em>

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