Identifying parts of a parabola from a graph, including vertex, axis of symmetry, intercepts, and solutions.

It looks like you're working on identifying key features of a quadratic function (parabola) from its graph. The image shows two parabolas, but the second one is incomplete with blanks to fill in. Let's go step by step and solve for the second parabola based on the information provided.

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Given for the Second Parabola:

- Vertex: $(-2, 4)$
- Axis of Symmetry: $x = -2$
- y-intercept: $(0, 4)$
- We need to find:
- x-intercepts
- Solutions (roots)

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Step 1: Understand the Graph

The vertex is at $(-2, 4)$, which means the highest point of the parabola is at $x = -2$, $y = 4$. Since the parabola opens downward (as seen from the shape), this is a maximum point.

The axis of symmetry is $x = -2$, which passes through the vertex.

The y-intercept is $(0, 4)$ — that’s where the graph crosses the y-axis.

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Step 2: Use Symmetry to Find x-intercepts

We know:

- The axis of symmetry is $x = -2$.
- The y-intercept is at $(0, 4)$, which is 2 units to the right of the axis of symmetry ($0 - (-2) = 2$).
- Because of symmetry, there should be a corresponding point 2 units to the left of the axis of symmetry at $x = -4$.

So, the point symmetric to $(0, 4)$ across $x = -2$ is $(-4, 4)$.

But we’re looking for x-intercepts, i.e., points where $y = 0$.

Let’s assume the parabola has two x-intercepts. Since it's symmetric about $x = -2$, if one x-intercept is at $x = a$, then the other must be at $x = -4 - a$? Wait — better to use the distance from the axis.

Let’s suppose the parabola crosses the x-axis at two points equidistant from $x = -2$.

Let’s denote the x-intercepts as:
$$
x = -2 + d \quad \text{and} \quad x = -2 - d
$$
for some $d > 0$

We don’t have exact values yet, so let’s try to infer from the graph.

Looking at the graph:
- The parabola appears to cross the x-axis at two points.
- It opens downward, vertex at $(-2, 4)$, and y-intercept at $(0, 4)$.
- So at $x = 0$, $y = 4$, and since the vertex is at $(-2, 4)$, both points $(-2, 4)$ and $(0, 4)$ are at the same height.

That suggests that the parabola is symmetric, and since $(-2, 4)$ is the vertex, and $(0, 4)$ is another point on the parabola, then the point symmetric to $(0, 4)$ across $x = -2$ is $(-4, 4)$.

So the parabola passes through:
- $(-4, 4)$
- $(-2, 4)$ ← vertex
- $(0, 4)$

Wait — that can't be unless the parabola is flat, but clearly it's curving down.

Wait — actually, the vertex is at $(-2, 4)$, and the y-intercept is at $(0, 4)$ — so at $x = 0$, $y = 4$, same as the vertex.

But if the vertex is the maximum, and the parabola goes downward, how can it be at the same height at $x = 0$?

Wait — that suggests the parabola is flat at $y = 4$ between $x = -4$ and $x = 0$? That doesn’t make sense unless it's not a standard parabola.

Wait — perhaps I made an error.

Let’s look again.

Wait — the y-intercept is given as $(0, 4)$, and the vertex is $(-2, 4)$.

So both the vertex and the y-intercept have the same y-value: $y = 4$.

Since the vertex is the highest point, and the parabola opens downward, then the point $(0, 4)$ is on the parabola, and at the same height as the vertex.

That means the parabola reaches $y = 4$ at $x = -2$ and again at $x = 0$.

But that would mean the parabola is not strictly decreasing after the vertex — unless it's symmetric.

Wait — if the vertex is at $x = -2$, and the parabola passes through $(0, 4)$, then the point symmetric to $(0, 4)$ across $x = -2$ is:

Distance from $x = -2$ to $x = 0$ is $+2$, so symmetric point is $-2 - 2 = -4$

So the parabola also passes through $(-4, 4)$

So far, we have:
- Vertex: $(-2, 4)$
- Points: $(-4, 4)$, $(-2, 4)$, $(0, 4)$

But that would mean the parabola is horizontal at $y = 4$ over $[-4, 0]$, which is impossible unless it's a line — but it's a parabola.

Wait — this can’t be right.

Wait — maybe I misread the y-intercept.

Let me double-check the image.

You wrote:

> y-intercept: (0, 4)

But wait — the vertex is at $(-2, 4)$, and the y-intercept is at $(0, 4)$ — both at $y = 4$.

But the graph shows the parabola going down from the vertex, so it should be lower than 4 at $x = 0$.

But you labeled the y-intercept as $(0, 4)$ — but visually, it seems like the y-intercept is below the vertex.

Wait — let's re-express.

Wait — in your image, the second parabola is drawn, and you’ve filled in:

- Vertex: $(-2, 4)$
- Axis of symmetry: $x = -2$
- y-intercept: $(0, 4)$
- x-intercepts: blank
- Solutions: blank

But looking at the graph, does the parabola pass through $(0, 4)$?

Let’s check the grid.

Assuming the grid lines are spaced by 1 unit.

From the graph:
- The vertex is at $(-2, 4)$ — correct.
- The y-axis is at $x = 0$.
- At $x = 0$, the graph appears to be at $y = 3$ or $y = 2$?

Wait — let’s carefully interpret the graph.

Looking at the blue parabola on the right:

- It peaks at $(-2, 4)$ — so vertex is correct.
- It crosses the y-axis (at $x = 0$) at $y = 3$ — because from the graph, it looks like it's one unit below the vertex.

But you wrote: y-intercept: $(0, 4)$ — that might be a mistake.

Wait — no, you wrote:

> y-intercept: (0, 4)

But visually, the graph at $x = 0$ is not at $y = 4$. It's lower.

Wait — perhaps you meant $(0, 3)$?

But in the first parabola, the y-intercept was correctly labeled as $(0, 3)$.

In the second parabola, the graph appears to go through $(0, 3)$, not $(0, 4)$.

But you wrote $(0, 4)$.

Let’s resolve this.

Wait — the vertex is at $(-2, 4)$, and the y-intercept is at $x = 0$.

On the graph, at $x = 0$, the point is not at $y = 4$. It's at $y = 3$.

So likely, the y-intercept is $(0, 3)$, not $(0, 4)$.

But you wrote $(0, 4)$ — possibly a typo.

Alternatively, maybe the graph is scaled differently.

Wait — let’s count the grid.

Suppose each square is 1 unit.

- The vertex is at $(-2, 4)$ — so at $x = -2$, $y = 4$
- The y-axis is at $x = 0$
- At $x = 0$, the curve is at $y = 3$ — yes, because it's one unit down from the vertex.

So y-intercept is $(0, 3)$, not $(0, 4)$.

But you wrote $(0, 4)$ — so that’s incorrect.

Let’s assume the correct y-intercept is $(0, 3)$.

Then:

- Vertex: $(-2, 4)$
- y-intercept: $(0, 3)$
- Axis of symmetry: $x = -2$

Now, let’s find the x-intercepts.

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Step 3: Use the vertex form of a parabola

The vertex form is:

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

where $(h, k)$ is the vertex.

Here, $h = -2$, $k = 4$, so:

$$
y = a(x + 2)^2 + 4
$$

Now plug in the y-intercept $(0, 3)$:

$$
3 = a(0 + 2)^2 + 4 \\
3 = a(4) + 4 \\
3 - 4 = 4a \\
-1 = 4a \\
a = -\frac{1}{4}
$$

So the equation is:

$$
y = -\frac{1}{4}(x + 2)^2 + 4
$$

Now, find x-intercepts by setting $y = 0$:

$$
0 = -\frac{1}{4}(x + 2)^2 + 4 \\
\frac{1}{4}(x + 2)^2 = 4 \\
(x + 2)^2 = 16 \\
x + 2 = \pm 4 \\
x = -2 \pm 4
$$

So:
- $x = -2 + 4 = 2$
- $x = -2 - 4 = -6$

Thus, the x-intercepts are:
- $(-6, 0)$ and $(2, 0)$

And the solutions (roots) are:
- $x = -6$ and $x = 2$

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Final Answers for the Second Parabola:

- Vertex: $(-2, 4)$ ✔
- Axis of Symmetry: $x = -2$ ✔
- y-intercept: $(0, 3)$ ✘ (you wrote $(0, 4)$ — that’s wrong)
- x-intercepts: $(-6, 0)$ and $(2, 0)$
- Solutions: $x = -6$ and $x = 2$

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Correction:

You wrote y-intercept: (0, 4) — but from the graph, it should be (0, 3).

Let me confirm: at $x = 0$, the graph is one unit below the vertex at $y = 4$, so $y = 3$.

So the correct entries are:

> Vertex: $(-2, 4)$
> Axis of Symmetry: $x = -2$
> y-intercept: $(0, 3)$
> x-intercepts: $(-6, 0)$ and $(2, 0)$
> Solutions: $x = -6$ and $x = 2$

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Summary:

| Feature | Value |
|-----------------------|--------------------------------|
| Vertex | $(-2, 4)$ |
| Axis of Symmetry | $x = -2$ |
| y-intercept | $(0, 3)$ |
| x-intercepts | $(-6, 0)$ and $(2, 0)$ |
| Solutions | $x = -6$ and $x = 2$ |

✔ This matches the graph.

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Let me know if you want the equation or further explanation!