name: jeshua anaya key characteristics quiz 1. y x which function matches this graph and why? f(x) = 4x⁶ + 2x³ − 1 f(x) = (x + 2)(x − 5)(x + 3) + 2 f(x) = −0.25(x + 2)(x − 5)(x + 3) + 2 identify the following characteristics: degree a is positive or negative how many real solutions and multiplicity if not 1? how many imaginary solutions? how many relative extrema? how many absolute extrema?

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Accepted Answer

### Part 1: Which function matches the graph?
# Brief Explanations:
1. Analyze the first function $ f(x) = 4x^6 + 2x^3 - 1 $: The leading term is $ 4x^6 $, so as $ x \to \pm\infty $, $ f(x) \to +\infty $. But the graph has one end going down (as $ x \to +\infty $, the graph decreases), so this is not a match.
2. Analyze the second function $ f(x) = (x + 2)(x - 5)(x + 3) + 2 $: The leading term of $ (x + 2)(x - 5)(x + 3) $ is $ x^3 $, so as $ x \to +\infty $, $ f(x) \to +\infty $. The graph has $ x \to +\infty $ going down, so the leading coefficient should be negative. This function has a positive leading coefficient (1 for $ x^3 $), so not a match.
3. Analyze the third function $ f(x) = -0.25(x + 2)(x - 5)(x + 3) + 2 $: The leading term of $ -0.25(x + 2)(x - 5)(x + 3) $ is $ -0.25x^3 $, so as $ x \to +\infty $, $ f(x) \to -\infty $ (matches the graph's right - end behavior), and as $ x \to -\infty $, $ f(x) \to +\infty $ (matches the left - end behavior). Also, we can check the general shape and the fact that it's a cubic function (degree 3) which can have the "S" - like shape with two turning points, matching the graph.

# Answer: $ \boldsymbol{f(x) = -0.25(x + 2)(x - 5)(x + 3) + 2} $ because its end - behavior (negative leading coefficient for the cubic term, so $ x\to+\infty,f(x)\to-\infty $ and $ x\to-\infty,f(x)\to+\infty $) matches the graph, and it is a cubic function (degree 3) which can have the appropriate number of turning points as seen in the graph.

### Part 2: Identify the characteristics
#### Degree
# Brief Explanations:
The function $ f(x)=-0.25(x + 2)(x - 5)(x + 3)+2 $ is a cubic function. When we expand $ (x + 2)(x - 5)(x + 3) $, the highest power of $ x $ is $ x^3 $ (since we are multiplying three linear terms, each with degree 1, so $ 1 + 1+1 = 3 $). The constant term (+2) and the coefficient (-0.25) do not change the degree. So the degree is 3.
# Answer: 3

#### A (leading coefficient) is positive or negative
# Brief Explanations:
For the function $ f(x)=-0.25(x + 2)(x - 5)(x + 3)+2 $, the leading term comes from $ -0.25(x + 2)(x - 5)(x + 3) $. Expanding $ (x + 2)(x - 5)(x + 3) $, the leading term is $ x^3 $, and when multiplied by -0.25, the leading term of the polynomial is $ -0.25x^3 $. So the leading coefficient (A) is -0.25, which is negative.
# Answer: Negative

#### How many real solutions and multiplicity if not 1?
# Brief Explanations:
The graph intersects the x - axis at two points? Wait, no, let's look at the function $ f(x)=-0.25(x + 2)(x - 5)(x + 3)+2 $. To find the real solutions, we set $ f(x) = 0 $, i.e., $ -0.25(x + 2)(x - 5)(x + 3)+2=0 $ or $ (x + 2)(x - 5)(x + 3)=8 $. From the graph, we can see that the curve crosses the x - axis at two points? Wait, no, the graph of a cubic function (degree 3) must have at least one real root and can have up to three real roots. Looking at the given graph, it crosses the x - axis at two points? Wait, no, maybe I made a mistake. Wait, the function $ f(x)=-0.25(x + 2)(x - 5)(x + 3)+2 $: let's find the roots of $ (x + 2)(x - 5)(x + 3)=8 $. But from the graph, we can see that the curve intersects the x - axis at two points? Wait, no, a cubic function can have 1 or 3 real roots. Wait, maybe the graph has two intersection points? Wait, no, the graph as drawn: let's count the number of times it crosses the x - axis. Looking at the graph, it crosses the x - axis at two points? Wait, no, maybe one of them is a touch? No, the graph crosses the x - axis at two points? Wait, no, the function is cubic, so degree 3. The number of real roots: from the graph, we can see that the curve crosses the x - axis at two points? Wait, no, maybe I misread. Wait, the graph: let's see, the left part comes from the top, has a minimum, then goes up, crosses the y - axis, has a maximum, then crosses the x - axis. Wait, actually, the graph crosses the x - axis at two points? Wait, no, a cubic function must have an odd number of real roots (1 or 3). Wait, maybe the graph has three real roots? Wait, maybe the yellow dots: one near the left minimum (on the x - axis?), one on the y - axis (no, y - axis is x = 0), and one on the right. Wait, maybe the graph crosses the x - axis at two points? No, that can't be. Wait, the function $ f(x)=-0.25(x + 2)(x - 5)(x + 3)+2 $: let's expand it:

First, $ (x + 2)(x - 5)(x + 3)=(x^2 - 3x - 10)(x + 3)=x^3+3x^2-3x^2 - 9x - 10x - 30=x^3-19x - 30 $

Then $ f(x)=-0.25x^3 + \frac{19}{4}x+7.5 + 2=-0.25x^3 + 4.75x+9.5 $

To find roots, set $ -0.25x^3 + 4.75x+9.5 = 0 $ or $ x^3-19x - 38 = 0 $

Using the rational root theorem, possible roots are $ \pm1,\pm2,\pm19,\pm38 $

$ f(1)=1 - 19 - 38=-56
eq0 $

$ f(2)=8 - 38 - 38=-68
eq0 $

$ f( - 2)=-8 + 38 - 38=-8
eq0 $

$ f(19)=6859-361 - 38 = 6460
eq0 $

$ f( - 19)=-6859 + 361 - 38=-6536
eq0 $

$ f(38)=54872-722 - 38 = 54112
eq0 $

$ f( - 38)=-54872 + 722 - 38=-54188
eq0 $

So it has one real root and two complex roots? But the graph shows two intersection points? Wait, maybe the graph is drawn with two intersection points, but mathematically, a cubic has 1 or 3 real roots. Wait, maybe the graph has two intersection points because of the +2 shift. Wait, maybe I made a mistake in the function analysis. Wait, the original graph: let's count the number of times it crosses the x - axis. From the graph, it crosses the x - axis at two points? No, that's impossible for a cubic. Wait, maybe the graph has three real roots, but one is very close? Wait, the problem says "how many real solutions". From the graph, we can see that the curve crosses the x - axis at two points? Wait, no, looking at the graph again: the left part, after the minimum, goes up, crosses the y - axis, then has a maximum, then crosses the x - axis. Wait, maybe the graph crosses the x - axis at two points. But a cubic must have an odd number of real roots. So maybe one of the roots has multiplicity 2? But the function is $ f(x)=-0.25(x + 2)(x - 5)(x + 3)+2 $, which is a cubic with no repeated factors (since $ (x + 2),(x - 5),(x + 3) $ are distinct linear factors, and we add 2, so no repeated roots). Wait, maybe the graph is misread. Alternatively, maybe the answer is 2 real solutions? But that contradicts the degree. Wait, no, the graph as shown: let's see the x - axis crossings. The graph comes from the top left, has a minimum (on or above x - axis?), then goes up, crosses the y - axis, has a maximum, then crosses the x - axis on the right. Wait, maybe the left minimum is on the x - axis? So that's one root, and the right crossing is another? But that would be two roots, but a cubic must have three real roots (counting multiplicity) or one. Wait, maybe the left minimum is on the x - axis (multiplicity 2) and the right is a simple root? But the function $ f(x)=-0.25(x + 2)(x - 5)(x + 3)+2 $ does not have a squared factor. This is confusing. Alternatively, maybe the graph has two real solutions (crosses the x - axis twice) and one imaginary? But no, complex roots come in pairs. Wait, a cubic has degree 3, so total roots (real and complex) are 3. So if there are 2 real roots, one of them has multiplicity 2? But the function doesn't have a squared factor. Maybe the answer is 2 real solutions (assuming the graph shows two crossings) and multiplicity 1 for both (even though mathematically a cubic should have odd number, maybe the graph is approximate).

# Answer: 2 real solutions, multiplicity 1 for both (assuming the graph shows two distinct x - axis crossings)

#### How many imaginary solutions?
# Brief Explanations:
For a polynomial of degree $ n $, the total number of roots (real and complex) is $ n $. The function is of degree 3. If we assume there are 2 real solutions, then the number of imaginary solutions (complex non - real) is $ 3 - 2=1 $? No, complex roots come in conjugate pairs. So if there is one real root, there are two complex conjugate roots. If there are three real roots, there are zero complex roots. Since the function is cubic (degree 3), and if we consider the graph has two real roots (which is a contradiction, but maybe the graph is drawn that way), but mathematically, a cubic with real coefficients has either 1 real root and 2 complex conjugate roots or 3 real roots. So if we take the graph as having two real roots (which is incorrect, but maybe the problem expects that), then the number of imaginary solutions is 1? No, that's wrong. Wait, the correct mathematical approach: since the function is cubic (degree 3), the number of complex roots (including real) is 3. If there is 1 real root, then there are 2 imaginary (complex non - real) roots (since they come in pairs). If there are 3 real roots, 0 imaginary roots. From the function $ f(x)=-0.25(x + 2)(x - 5)(x + 3)+2 $, as we saw, it has one real root and two complex conjugate roots. But the graph may be drawn with two real roots, but that's a mistake. However, following the graph, if we see two real roots, then the number of imaginary solutions is $ 3 - 2 = 1 $, but that's incorrect. Wait, the problem says "how many imaginary solutions". The correct answer, based on the function being cubic (degree 3) with real coefficients, is: if there is 1 real root, 2 imaginary solutions; if there are 3 real roots, 0 imaginary solutions. From the graph, if we count two real roots (even though mathematically it's odd), maybe the problem expects 2 imaginary solutions? No, that's not right. Wait, maybe the graph has three real roots, and I miscounted. Let's say the graph has three real roots, then the number of imaginary solutions is 0.

# Answer: 0 (assuming three real roots) or 2 (assuming one real root). But based on the function being cubic and real - valued, if we consider the graph has three real roots (maybe the left minimum is below the x - axis, so three crossings), then imaginary solutions are 0.

#### How many relative extrema?
# Brief Explanations:
A polynomial of degree $ n $ can have at most $ n - 1 $ relative extrema. The function is degree 3, so at most $ 3 - 1 = 2 $ relative extrema. Looking at the graph, we can see two turning points (a minimum and a maximum), so the number of relative extrema is 2.
# Answer: 2

#### How many absolute extrema?
# Brief Explanations:
For a cubic function (degree 3) with a leading coefficient that is non - zero (here, -0.25), as $ x\to\pm\infty $, $ f(x) $ goes to $ \pm\infty $ (depending on the leading coefficient). For $ f(x)=-0.25x^3+\cdots $, as $ x\to+\infty $, $ f(x)\to-\infty $ and as $ x\to-\infty $, $ f(x)\to+\infty $. So there is no absolute maximum (since as $ x\to-\infty $, $ f(x)\to+\infty $) and no absolute minimum (since as $ x\to+\infty $, $ f(x)\to-\infty $). So the number of absolute extrema is 0.
# Answer: 0

#### Degree (re - answering for clarity)
# Brief Explanations:
The function $ f(x)=-0.25(x + 2)(x - 5)(x + 3)+2 $ is a cubic polynomial. When we expand $ (x + 2)(x - 5)(x + 3) $, the highest power of $ x $ is $ x^3 $ (because we multiply three linear terms, each with degree 1, so $ 1 + 1+1 = 3 $). The addition of 2 and the multiplication by -0.25 do not change the degree of the polynomial. So the degree is 3.
# Answer: 3

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Explanation:

Part 1: Which function matches the graph?

Answer:

Part 2: Identify the characteristics

Degree