Question

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Answer 1

f(x) = x² +6x +5

Explanation:

You want the equation of the parabola with vertex (-3, -4) and y-intercept (0, 5) written in standard form.

Standard form

The standard form equation will look like ...

f(x) = ax² +bx +c

where c is the y-intercept.

Vertex form

The vertex of the parabola is the turning point, shown as (-3, -4). This can be used to write the parabola's equation in vertex form:

f(x) = a(x -h)² +k . . . . . . . . vertex (h, k) and vertical scale factor 'a'

Using the given vertex, we have ...

f(x) = a(x +3)² -4

The y-intercept is the value when x=0, the y-value where the graph crosses the y-axis.

f(0) = a(0 +3)² -4 = 9a -4 . . . . . . function value at x=0

We know this is 5, so ...

5 = 9a -4

9 = 9a . . . . . . . add 4

a = 1 . . . . . . . . divide by 9

Equation

Then the standard form equation can be found by expanding ...

f(x) = 1(x +3)² -4

f(x) = x² +6x +5²

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Additional comment

The vertex and y-intercept are marked on the graph, so this becomes a vocabulary question. You have to understand the meaning of the terms "vertex" and "y-intercept" and how to find them on a graph.

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Answer 2

Answer and Explanation:

The vertex of a parabola is the point where the curve changes direction.

From the graph, we can determine that this point is:

(-3, -4)

To get an equation for the parabola, we can plug this into the vertex form of a quadratic equation:

`y = m(x - a)^2 + b`

where
`(a,b)` is the parabola's vertex and
`m` is its vertical stretch relative to a standard parabola (
`y=(x-a)^2 + b`).

We can identify the following variable values:

• 
  `a = -3`
• 
  `b = -4`
• 
  `m=1` → this means the parabola is not stretched

Now, we can plug these values into the vertex form equation:

`y = 1(x - (-3))^2 + (-4)`

and simplify:

`\huge\boxed{y = (x + 3)^2 - 4}`