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3 years ago

Write the quadratic function in vertex form. y = x2 + 4x + 7

Mathematics

1 answer:

larisa86 [58]3 years ago

8 0

Complete the square on the right side of the equation.

Use the form ax^2 + bx + cax^2 + bx + c, to find the values of aa , bb, and cc.

a=1, b=4, c=7

Consider the vertex form of a parabola.

a(x + d)^2 + e

Find the value of d using the formula d = b / 2a

Multiply 2 by 1 to get 2 ⋅ 1.

d = 4 / (2 * 1)

d = 2

Find the value of e using the formula e =c −b^2 / 4a

Multiply 4 by 1 to get 4 ⋅ 1

E = 7 – ((4)^2 / (4 ⋅ 1))

Reduce the expression by cancelling the common factors.

E = 7 – 1 ⋅ 4

Subtract 44 from 77 to get 33.

e = 3

Substitute the values of a, d, and e into the vertex form a(x + d)^2 + e.

(x + 2)^2 + 3

Therefore, the vertex form of the quadratic equation is:

y = (x + 2)^2 + 3

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3 years ago

in a tidal river, the time between high and low tide is 7.6 hours. At high tide the depth of water is 17.2 feet, while at low ti

Lady_Fox [76]

Alright, so the difference between tidal heights is 12.6 ft. right? For every integer cos(x) is multiplied, say a*cos(x) the amplitude doubles. So then, (1/2)=(n/12.6) will gives us a value for a that will match the differences in tidal height. a=6.3. Now let's say that 2pi will give us a full cycle from high to low back to high again. If 12:00 noon is our high, it takes 15.2hrs to complete a cycle back to high again. 2pi = 6.28 roughly, and we know that is less than our midpoint in a whole cycle, 7.6 hrs. Also, for every value k>1 in cos(kx), this increases the frequency of waves in a given period. So it seems that we need a value of k less than 1. So we can lower the frequency to match the cycle we are looking for, which is to come back to 17.7 ft by 15.2hrs. So to find our value k we divide 2pi by 15.2. This is 0.41. We do it that way based on the ratio for the standard period to what we need, which is a number that can get us to 15.2. So far we have f(t) = 6.3cos(0.41x). Now the problem we have now is that we are not at our height for max, nor our min, even though our wave matches the difference between them, we are between 6.3 and -6.3 so far. so for every value of c in cos(x)+c, this adds to the height cos starts on the y-axis. Since we are at 6.3 already, all we need to do is subtract 6.3 from 17.6 to get c. So C is 11.3 and our final equation is f(t)=6.3cos(0.41x)+11.3. This gives our tides in 15.2 hr periods, starting at 12 noon = x=0, with extremes at intervals of 7.6 hours. Let me know if you have any more questions about it, it's kind of rough to explain. We transformed by f(t)=a*cos(kx)+c altogether, but I thought it might help to separate each value independently.

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3 years ago

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3 years ago

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2 years ago

to solve the system of equations below, Becca isolated x^2 in the firs equation and then substituted it into the second equation

bija089 [108]

The resulting equation if Becca isolated x² in the first equation and then substituted it into the second equation is (9-y²) / 25 - y²/36 = 1

<h3>Equation</h3>

x² + y² = 9

x²/25 - y²/36 = 1

From (1)

x² = 9 - y²

substitute x² = 9 - y² into (2)

x²/25 - y²/36 = 1

(9-y²) / 25 - y²/36 = 1

Therefore, the resulting equation is option C; (9-y²) / 25 - y²/36 = 1

Learn more about equation:

brainly.com/question/4172455

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2 years ago