Standard Form Practice worksheet with three quadratic function problems involving graph analysis and graphing.

Let's solve each problem step by step using the given information and the graphs provided.

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Problem 1:

Function: $ f(x) = x^2 + 4x + 7 $

We are to analyze the graph of this quadratic function. Let’s find its characteristics.

#### Step 1: Domain
- The domain of a quadratic function (a parabola) is all real numbers because there are no restrictions on $ x $.
- Domain: $ (-\infty, \infty) $

#### Step 2: Range
- Since the coefficient of $ x^2 $ is positive ($ a = 1 > 0 $), the parabola opens upward, so it has a minimum value at the vertex.
- From the graph, we can see the lowest point (vertex) is around $ y = 3 $.
- So the range starts from that minimum and goes up to infinity.
- Range: $ [3, \infty) $

#### Step 3: Axis of Symmetry
- For a quadratic in standard form $ f(x) = ax^2 + bx + c $, the axis of symmetry is:
$$
x = -\frac{b}{2a}
$$
Here, $ a = 1 $, $ b = 4 $:
$$
x = -\frac{4}{2(1)} = -2
$$
- Axis of Symmetry: $ x = -2 $

#### Step 4: Vertex
- The vertex lies on the axis of symmetry. Plug $ x = -2 $ into the function:
$$
f(-2) = (-2)^2 + 4(-2) + 7 = 4 - 8 + 7 = 3
$$
- So, Vertex: $ (-2, 3) $

#### Step 5: Y-intercept
- This occurs when $ x = 0 $:
$$
f(0) = 0^2 + 4(0) + 7 = 7
$$
- Y-intercept: $ (0, 7) $

#### Step 6: Zero(s)
- These are the $ x $-intercepts (where $ f(x) = 0 $).
- Look at the graph: does the parabola cross the $ x $-axis?
- The vertex is at $ y = 3 $, and it opens upward → it never touches or crosses the $ x $-axis.
- So, no real zeros.
- Zeros: None (or "No real solutions")

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✔ Answer for Problem 1:

- Domain: $ (-\infty, \infty) $
- Range: $ [3, \infty) $
- Axis of Symmetry: $ x = -2 $
- Vertex: $ (-2, 3) $
- Y-intercept: $ (0, 7) $
- Zero(s): None

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Problem 2:

Function: $ y = -x^2 - 5x + 1 $

Again, use the graph to find the characteristics.

#### Step 1: Domain
- Again, this is a quadratic — domain is all real numbers.
- Domain: $ (-\infty, \infty) $

#### Step 2: Range
- Coefficient of $ x^2 $ is negative ($ a = -1 < 0 $), so the parabola opens downward, meaning it has a maximum at the vertex.
- From the graph, the highest point (vertex) appears to be around $ y = 7.25 $.
- But let's compute it exactly.

Use formula:
$$
x = -\frac{b}{2a} = -\frac{-5}{2(-1)} = -\frac{5}{2} = -2.5
$$
Now plug into function:
$$
y = -(-2.5)^2 - 5(-2.5) + 1 = -6.25 + 12.5 + 1 = 7.25
$$
So maximum $ y = 7.25 = \frac{29}{4} $

Thus, Range: $ (-\infty, 7.25] $ or $ (-\infty, \frac{29}{4}] $

#### Step 3: Axis of Symmetry
- As computed: $ x = -2.5 $
- Axis of Symmetry: $ x = -\frac{5}{2} $ or $ x = -2.5 $

#### Step 4: Vertex
- $ (-2.5, 7.25) $ or $ \left(-\frac{5}{2}, \frac{29}{4}\right) $
- Vertex: $ \left(-\frac{5}{2}, \frac{29}{4}\right) $

#### Step 5: Y-intercept
- Set $ x = 0 $:
$$
y = -(0)^2 - 5(0) + 1 = 1
$$
- Y-intercept: $ (0, 1) $

#### Step 6: Zero(s)
- Find where the graph crosses the $ x $-axis.
- From the graph, it crosses at two points:
- One between $ x = -6 $ and $ x = -5 $
- One between $ x = 0 $ and $ x = 1 $
- We can estimate:
- Left zero: about $ x = -5.2 $
- Right zero: about $ x = 0.2 $
- But let’s solve algebraically:

Set $ -x^2 - 5x + 1 = 0 $

Multiply both sides by $-1$:
$ x^2 + 5x - 1 = 0 $

Use quadratic formula:
$$
x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-1)}}{2(1)} = \frac{-5 \pm \sqrt{25 + 4}}{2} = \frac{-5 \pm \sqrt{29}}{2}
$$

So,
$$
x = \frac{-5 + \sqrt{29}}{2} \approx \frac{-5 + 5.385}{2} \approx \frac{0.385}{2} \approx 0.1925
$$
$$
x = \frac{-5 - \sqrt{29}}{2} \approx \frac{-5 - 5.385}{2} \approx \frac{-10.385}{2} \approx -5.1925
$$

So Zeros: $ x = \frac{-5 \pm \sqrt{29}}{2} $

But since the question says "use the graph", approximate values from the graph are acceptable.

From graph:
- Zeros ≈ $ x = -5.2 $ and $ x = 0.2 $

But better to write exact form if possible.

However, based on the graph grid, we can estimate:
- One zero at $ x = -5.2 $, other at $ x = 0.2 $

But since the graph may not be perfectly accurate, let’s go with approximate decimal values.

Zeros: $ x \approx -5.2 $ and $ x \approx 0.2 $

Or more precisely: $ x \approx -5.19 $ and $ x \approx 0.19 $

But for simplicity, use estimated values from graph.

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✔ Answer for Problem 2:

- Domain: $ (-\infty, \infty) $
- Range: $ (-\infty, 7.25] $ or $ (-\infty, \frac{29}{4}] $
- Axis of Symmetry: $ x = -2.5 $
- Vertex: $ (-2.5, 7.25) $ or $ \left(-\frac{5}{2}, \frac{29}{4}\right) $
- Y-intercept: $ (0, 1) $
- Zero(s): $ x \approx -5.2 $ and $ x \approx 0.2 $ (from graph)

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Problem 3: Graph the quadratic function

Function: $ f(x) = 2x^2 + x - 6 $

We need to graph this quadratic. Let’s find key features.

#### Step 1: Direction of Opening
- Coefficient of $ x^2 $ is $ 2 > 0 $ → opens upward

#### Step 2: Vertex
Use:
$$
x = -\frac{b}{2a} = -\frac{1}{2(2)} = -\frac{1}{4} = -0.25
$$
Now find $ f(-0.25) $:
$$
f(-0.25) = 2(-0.25)^2 + (-0.25) - 6 = 2(0.0625) - 0.25 - 6 = 0.125 - 0.25 - 6 = -6.125
$$
So vertex is at $ (-0.25, -6.125) $

#### Step 3: Y-intercept
Set $ x = 0 $:
$$
f(0) = 2(0)^2 + 0 - 6 = -6
$$
→ $ (0, -6) $

#### Step 4: X-intercepts (Zeros)
Solve $ 2x^2 + x - 6 = 0 $

Use quadratic formula:
$$
x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-6)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 48}}{4} = \frac{-1 \pm \sqrt{49}}{4} = \frac{-1 \pm 7}{4}
$$

So:
- $ x = \frac{-1 + 7}{4} = \frac{6}{4} = 1.5 $
- $ x = \frac{-1 - 7}{4} = \frac{-8}{4} = -2 $

Zeros: $ x = -2 $, $ x = 1.5 $

#### Step 5: Additional Points
To graph, plot:
- Vertex: $ (-0.25, -6.125) $
- Y-intercept: $ (0, -6) $
- X-intercepts: $ (-2, 0) $, $ (1.5, 0) $
- Try $ x = 1 $: $ f(1) = 2(1)^2 + 1 - 6 = 2 + 1 - 6 = -3 $
- Try $ x = -1 $: $ f(-1) = 2(1) -1 -6 = 2 -1 -6 = -5 $
- Try $ x = 2 $: $ f(2) = 2(4) + 2 - 6 = 8 + 2 - 6 = 4 $

Plot these points and draw a smooth parabola opening upward.

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✔ Graphing Steps Summary:

1. Plot the vertex: $ (-0.25, -6.125) $
2. Plot x-intercepts: $ (-2, 0) $, $ (1.5, 0) $
3. Plot y-intercept: $ (0, -6) $
4. Add extra points: $ (-1, -5) $, $ (1, -3) $, $ (2, 4) $
5. Draw a smooth U-shaped curve through them.

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Final Answer Summary:

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Problem 1:

- Domain: $ (-\infty, \infty) $
- Range: $ [3, \infty) $
- Axis of Symmetry: $ x = -2 $
- Vertex: $ (-2, 3) $
- Y-intercept: $ (0, 7) $
- Zero(s): None

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Problem 2:

- Domain: $ (-\infty, \infty) $
- Range: $ (-\infty, 7.25] $
- Axis of Symmetry: $ x = -2.5 $
- Vertex: $ (-2.5, 7.25) $
- Y-intercept: $ (0, 1) $
- Zero(s): $ x \approx -5.2 $, $ x \approx 0.2 $

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Problem 3: Graph of $ f(x) = 2x^2 + x - 6 $

- Opens upward
- Vertex: $ (-0.25, -6.125) $
- Y-intercept: $ (0, -6) $
- X-intercepts: $ (-2, 0) $, $ (1.5, 0) $
- Additional points: $ (-1, -5) $, $ (1, -3) $, $ (2, 4) $

> Graph: Plot these points and draw a parabola opening upward passing through them.

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Let me know if you'd like a visual sketch description or help plotting on the grid!