Keys on a keyboard because it can’t be letters because there are numbers and punctuation marks, it can’t be numbers because there are letters and punctuation marks, it can’t be punctuation merks either cause there are letters and numbers. So the only option left is keys on a keyboard cause it has all letters numbers and punctuation marks

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1 year ago

How do I solve question 6 through 8?<br> Solve for me

rewona [7]

The equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2

<h3>How to determine the functions?</h3>

A quadratic function is represented as:

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

<u>Question #6</u>

The vertex of the graph is

(h, k) = (-1, 2)

So, we have:

y = a(x + 1)^2 + 2

The graph pass through the f(0) = -2

So, we have:

-2 = a(0 + 1)^2 + 2

Evaluate the like terms

a = -4

Substitute a = -4 in y = a(x + 1)^2 + 2

y = -4(x + 1)^2 + 2

<u>Question #7</u>

The vertex of the graph is

(h, k) = (2, 1)

So, we have:

y = a(x - 2)^2 + 1

The graph pass through (1, 3)

So, we have:

3 = a(1 - 2)^2 + 1

Evaluate the like terms

a = 2

Substitute a = 2 in y = a(x - 2)^2 + 1

y = 2(x - 2)^2 + 1

<u>Question #8</u>

The vertex of the graph is

(h, k) = (1, -2)

So, we have:

y = a(x - 1)^2 - 2

The graph pass through (0, -3)

So, we have:

-3 = a(0 - 1)^2 - 2

Evaluate the like terms

a = -1

Substitute a = -1 in y = a(x - 1)^2 - 2

y = -(x - 1)^2 - 2

Hence, the equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2