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4 years ago

Write the quadratic function in vertex form. Y = x2 + 16x + 74

Mathematics

1 answer:

Sidana [21]4 years ago

8 0

<u>Answer-</u>

The vertex form of the given quadratic function is,

$\boxed {\boxed {y = (x+8)^2+10}}$

<u>Solution-</u>

The equation for a parabola or quadratic function can be written in <em>vertex form</em>-

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

The given quadratic function,

$\Rightarrow y = x^2 + 16x + 74$

$\Rightarrow y = (x)^2 + (2\times x \times \frac{16}{2}) + 74$

$\Rightarrow y = (x)^2 + (2\times x \times 8) + 74$

$\Rightarrow y = (x)^2 + (2\times x \times 8) + (8)^2+74-8^2$

$\Rightarrow y = (x+8)^2+74-64$

$\Rightarrow y = (x+8)^2+10$

This is the vertex form of the given quadratic function with vertex at (-8, 10)

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You can find the answer to any quadratic by using the quadratic formula. The formula is below.

$\frac{-b +/- \sqrt{b^{2}-4ac}}{2a}$

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$\frac{2 +/- \sqrt{-2^{2}-4(6)(-7)}}{2(6)}$

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3 years ago

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3 years ago

1. Find four consecutive even integers such that the sum of the first and third is 6 less than the largest

Sophie [7]

These are indeed quite a lot of exercises, but a lot of them are almost identical - only small computations will change. So, I'm glad to help you with one exercise from every cathegory, but I encourage you to solve the others on your own.

<h2>Exercise 1</h2>

You can call four consecutive integers as

$x,\ x+1,\ x+2,\ x+3$

So, the sum of the first and third is $x+(x+2) = 2x+2$

We want this quantity to be six less than the largest, i.e. $(x+3)-6 = x-3$

So, the equality is

$2x+2 = x-3$

Subtract x from both sides:

$x+2 = -3$

Subtract 2 from both sides:

$x = -5$

So, the consecutive integers are

$-5,\ -4,\ -3,\ -2$

In fact, the sum of the first and third is $-5-3 = -8$ , which is indeed six less than the largest: $-8 = -2-6$

<h2>Exercise 2</h2>

If you call the first odd number $x$ , the next consecutive odd numbers will be $x,\ x+2,\ x+4,\ x+6$

In fact, we have to count skipping two's, because we only want odd integers. From here, you go on like exercise 1: you write the largest (which is x+6), and set it to be two more than the sum of the other three (x, x+2 and x+4)

<h2>Exercise 3</h2>

By the same logic of exercise 2, two consecutive even integers are $x,\ x+2$ , assuming that x is even.

So, you set the equation as usual: the smaller (which is x) is 26 less than three times the larger (which means 3(x+2)-26)

<h2>Exercise 4 to 8</h2>

These are all pretty identical to exercise 1: you start by listing three or four consecutive integers:

$x,\ x+1,\ x+2\quad\text{or}\quad x,\ x+1,\ x+2\ x+3$

and then you translate the request of each exercise accordingly. Remember that expressions like "three times the second number" means that you have to multiply: $3(x+1)$ , while expression like "six more than the first" or "thirteen less than the first" imply adding/subtracting: $x+6$ or $x-13$ .