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3 years ago

F(x) = 2x^2 − 3x − 5 Part

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

5 0

Answer: Answer is in the steps..

Step-by-step explanation:

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

A) To find the x intercepts of a parabola we have to set the whole equation equal zero because the x intercept is when y equal zero in other words where f(x) equal zero.

$2x^{2}$ - 3x - 5 = 0 Now solve using quadratic formula.

x = -b± $\frac{\sqrt{b^2 -4ac } }{2a}$

x = 3 ± $\sqrt{9+40} /2a$

x = 3 ± $\sqrt{49}$ / 2a

x= 3 ± 7 /4

x= -1 or x = 5/2

The x intercepts are (5/2,0) and (-1,0)

B) The graph vertex of the graph of f(x) is going to be at a minimum because the leading coefficient of the function has a positive integer which means the parabola will open up and its vertex will be at the minimum point.

Using the x coordinates we could find their average and find the x coordinate of the vertex.

$\frac{5}{2} + -\frac{1}{1}$ = $\frac{3}{2} / \frac{2}{1}$ = 3/4

The x coordinate of the vertex is 3/4 so we will input that into the equation and solve for the y coordinate

$2(\frac{3}{4})^2 - 3 (\frac{3}{4}) - 5$ = 18/16 - 9/4 -5 = -49/8

The y coordinate of the vertex is -49/8

So the vertex is (3/4, -49/8)

C) First I will graph the vertex and then graph he two x intercepts and connect then like a U shape.

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3 years ago

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3 years ago

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3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

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2 years ago