which function matches this graph and why?
$f(x) = -2x^6 - 13x^5 + 20x$
$f(x) = 2x^6 - 13x^5 + 26x^4 - 7x^3 - 28x^2 + 20x$
$f(x) = 2x(x + 7)(x - 4)(x + 3)(x - 2) - 3$
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? ________

Question

Accepted Answer

### Part 1: Which function matches the graph?
# Brief Explanations:
1. Analyze the leading term: The graph has both ends going up (as $ x \to \pm\infty $, $ y \to +\infty $), so the leading coefficient must be positive and the degree even. The first function $ f(x) = -2x^6 - 13x^5 + 20x $ has a negative leading coefficient ($-2$) and degree 6 (even), but ends would go down, so eliminate. The third function $ f(x) = 2x(x + 7)(x - 4)(x + 3)(x - 2) - 3 $ is a 5th-degree polynomial (odd degree), so ends go in opposite directions, eliminate. The second function $ f(x) = 2x^6 - 13x^5 + 26x^4 - 7x^3 - 28x^2 + 20x $ has a positive leading coefficient ($2$) and degree 6 (even), matching the end behavior. Also, factor it: $ f(x) = x(2x^5 - 13x^4 + 26x^3 - 7x^2 - 28x + 20) $, and further factoring shows real roots, consistent with the graph’s x-intercepts.
# Answer:
$ \boldsymbol{f(x) = 2x^6 - 13x^5 + 26x^4 - 7x^3 - 28x^2 + 20x} $ (because it has a positive leading coefficient, even degree (6) matching end behavior, and factors to show real roots consistent with the graph’s x - intercepts, while the other functions fail end - behavior or degree checks).

### Part 2: Identify the characteristics (for the correct function $ f(x) = 2x^6 - 13x^5 + 26x^4 - 7x^3 - 28x^2 + 20x $)
#### Degree:
# Brief Explanations:
The highest power of $ x $ in the polynomial is $ x^6 $, so the degree is 6.
# Answer:
$ \boldsymbol{6} $

#### A (leading coefficient) is positive or negative:
# Brief Explanations:
The leading term is $ 2x^6 $, so the leading coefficient $ A = 2 $, which is positive.
# Answer:
$ \boldsymbol{\text{Positive}} $

#### Real solutions (roots) and multiplicity:
# Brief Explanations:
Factor $ f(x) = x(2x^5 - 13x^4 + 26x^3 - 7x^2 - 28x + 20) $. The root $ x = 0 $ has multiplicity 1. The 5th - degree polynomial factor can be factored further (e.g., by rational root theorem) to find other real roots. From the graph, we see multiple x - intercepts. Let's count: the graph crosses the x - axis at least 4 times (visually), and with $ x = 0 $, total real roots: let's check the factored form or graph. The polynomial is degree 6, so total roots (real + imaginary) are 6. From the graph, we can see 4 real roots (including $ x = 0 $ and others) with multiplicity 1 (since the graph crosses the axis at each, not touches, so no even multiplicity). Wait, actually, factoring $ f(x)=x(2x - 1)(x - 1)(x - 2)(x - 4)(x + 1) $? Wait, no, let's re - factor: $ f(x)=2x^6 - 13x^5 + 26x^4 - 7x^3 - 28x^2 + 20x=x(2x^5 - 13x^4 + 26x^3 - 7x^2 - 28x + 20) $. Using rational root theorem, possible roots are $ \pm1,\pm2,\pm4,\pm5,\pm10,\pm20,\pm\frac{1}{2},\pm\frac{5}{2} $. Testing $ x = 1 $: $ 2 - 13 + 26 - 7 - 28 + 20 = 0 $, so $ (x - 1) $ is a factor. Dividing the 5th - degree polynomial by $ (x - 1) $ gives $ 2x^4 - 11x^3 + 15x^2 + 8x - 20 $. Testing $ x = 2 $: $ 32 - 88 + 60 + 16 - 20 = 0 $, so $ (x - 2) $ is a factor. Dividing by $ (x - 2) $ gives $ 2x^3 - 7x^2 + x + 10 $. Testing $ x = - 1 $: $ - 2 - 7 - 1 + 10 = 0 $, so $ (x + 1) $ is a factor. Dividing by $ (x + 1) $ gives $ 2x^2 - 9x + 10 $, which factors to $ (2x - 5)(x - 2) $. Wait, so $ f(x)=x(x - 1)(x - 2)(x + 1)(2x - 5)(x - 2) $? Wait, no, let's do it correctly. Wait, original function: $ f(x)=2x^6 - 13x^5 + 26x^4 - 7x^3 - 28x^2 + 20x $. Let's factor step by step:

$ f(x)=x(2x^5 - 13x^4 + 26x^3 - 7x^2 - 28x + 20) $

Let $ g(x)=2x^5 - 13x^4 + 26x^3 - 7x^2 - 28x + 20 $

$ g(1)=2 - 13 + 26 - 7 - 28 + 20 = 0 $, so $ (x - 1) $ is a factor.

Divide $ g(x) $ by $ (x - 1) $: Using polynomial long division or synthetic division, we get $ 2x^4 - 11x^3 + 15x^2 + 8x - 20 $

$ g_1(x)=2x^4 - 11x^3 + 15x^2 + 8x - 20 $

$ g_1(2)=32 - 88 + 60 + 16 - 20 = 0 $, so $ (x - 2) $ is a factor.

Divide $ g_1(x) $ by $ (x - 2) $: $ 2x^3 - 7x^2 + x + 10 $

$ g_2(x)=2x^3 - 7x^2 + x + 10 $

$ g_2(-1)=-2 - 7 - 1 + 10 = 0 $, so $ (x + 1) $ is a factor.

Divide $ g_2(x) $ by $ (x + 1) $: $ 2x^2 - 9x + 10 $

Factor $ 2x^2 - 9x + 10=(2x - 5)(x - 2) $

So $ g(x)=(x - 1)(x - 2)(x + 1)(2x - 5)(x - 2) $? Wait, no, when we divided $ g_2(x) $ by $ (x + 1) $, we get $ 2x^2 - 9x + 10 $, so $ g(x)=(x - 1)(x - 2)(x + 1)(2x^2 - 9x + 10)=(x - 1)(x - 2)(x + 1)(2x - 5)(x - 2) $? Wait, $ 2x^2 - 9x + 10=(2x - 5)(x - 2) $, so $ g(x)=(x - 1)(x - 2)^2(x + 1)(2x - 5) $

Thus, $ f(x)=x(x - 1)(x - 2)^2(x + 1)(2x - 5) $

So real roots: $ x = 0 $ (multiplicity 1), $ x = 1 $ (multiplicity 1), $ x = 2 $ (multiplicity 2), $ x=-1 $ (multiplicity 1), $ x=\frac{5}{2} $ (multiplicity 1). Wait, but from the graph, how many real solutions? Let's count the x - intercepts. The graph crosses the x - axis at 5 points? Wait, no, the original graph: let's see, the function $ f(x)=2x^6 - 13x^5 + 26x^4 - 7x^3 - 28x^2 + 20x $ has roots at $ x = 0 $, $ x = 1 $, $ x = 2 $ (with multiplicity 2, so the graph touches the axis at $ x = 2 $ and turns around), $ x=-1 $, $ x=\frac{5}{2} $. So real solutions: 5 distinct real roots? Wait, no, $ x = 2 $ has multiplicity 2, so total real roots (counting multiplicity) is $ 1 + 1+2 + 1+1 = 6 $, but distinct real roots: $ x = 0,1,2,-1,\frac{5}{2} $ (5 distinct, with $ x = 2 $ having multiplicity 2). But from the graph, visually, we can see that there are 4 or 5 x - intercepts. Wait, maybe my factoring is wrong. Alternatively, from the graph, we can see that the function crosses the x - axis at least 4 times, and has a touch at one point (multiplicity 2). But let's go back to the graph: the graph has a local maximum, then a minimum, then crosses, then a touch (where it touches and turns, multiplicity 2), then crosses again. So real solutions: Let's count the number of times it intersects the x - axis. From the graph, we can see 5 real solutions? Wait, no, the degree is 6, so total roots (real + imaginary) is 6. If we have a root with multiplicity 2 (touching the axis), then the number of distinct real roots: let's see, from the factored form we had $ f(x)=x(x - 1)(x - 2)^2(x + 1)(2x - 5) $, so real roots: $ x = 0,1,2,-1,\frac{5}{2} $ (5 distinct, with $ x = 2 $ having multiplicity 2). So number of real solutions (distinct) with multiplicity: $ x = 0 $ (1), $ x = 1 $ (1), $ x = 2 $ (2), $ x=-1 $ (1), $ x=\frac{5}{2} $ (1). But maybe the graph shows 4 distinct x - intercepts (with one being a touch). Anyway, the key is: degree 6, so total real roots (counting multiplicity) is 6. But from the graph, let's assume that we can see 5 real solutions (with one having multiplicity 2). Wait, maybe I made a mistake in factoring. Let's use the graph: the function is a 6th - degree polynomial. The end behavior is up - up. The graph has, say, 5 x - intercepts (one with multiplicity 2). So number of real solutions: Let's say 5 real solutions (with one having multiplicity 2). But maybe the correct way is: from the graph, we can see that the function crosses the x - axis at 4 points and touches at 1 point, so total real roots (counting multiplicity) is 6 (since degree 6). So number of real solutions (distinct) with multiplicity: Let's say 5 real solutions (one with multiplicity 2). But maybe the answer expects: real solutions: 5 (with $ x = 2 $ having multiplicity 2), or based on the graph, we can see that there are 5 real x - intercepts (including the touch).

# Brief Explanations:
From the factored form $ f(x)=x(x - 1)(x - 2)^2(x + 1)(2x - 5) $, the real roots are $ x = 0 $ (multiplicity 1), $ x = 1 $ (multiplicity 1), $ x = 2 $ (multiplicity 2), $ x=-1 $ (multiplicity 1), $ x=\frac{5}{2} $ (multiplicity 1). So there are 5 distinct real solutions, with $ x = 2 $ having multiplicity 2, and the rest having multiplicity 1.
# Answer:
$ \boldsymbol{5 \text{ distinct real solutions: } x = 0 \text{ (multiplicity 1)}, x = 1 \text{ (multiplicity 1)}, x = 2 \text{ (multiplicity 2)}, x=-1 \text{ (multiplicity 1)}, x=\frac{5}{2} \text{ (multiplicity 1)}} $ (or, if counting total real roots with multiplicity: 6, with $ x = 2 $ having multiplicity 2 and others multiplicity 1)

#### Imaginary solutions:
# Brief Explanations:
A degree 6 polynomial has 6 total roots (real + imaginary). From the factored form, we have 6 real roots (counting multiplicity: $ 1 + 1+2 + 1+1 = 6 $), so there are 0 imaginary solutions.
# Answer:
$ \boldsymbol{0} $

#### Relative extrema:
# Brief Explanations:
A degree $ n $ polynomial can have at most $ n - 1 $ relative extrema. For $ n = 6 $, at most 5 relative extrema. From the graph, we can see peaks and valleys: let's count the number of times the slope changes from increasing to decreasing or vice - versa. The graph has a local minimum, a local maximum, a local minimum, a local maximum, and a local minimum? Wait, looking at the graph, we can see 5 relative extrema (4 minima and 1 maximum? Or 3 minima and 2 maxima? Wait, the function $ f(x)=2x^6 - 13x^5 + 26x^4 - 7x^3 - 28x^2 + 20x $ has derivative $ f^\prime(x)=12x^5 - 65x^4 + 104x^3 - 21x^2 - 56x + 20 $. The number of relative extrema is the number of real roots of $ f^\prime(x) $. From the graph, we can see 5 relative extrema (since degree 6, at most 5).
# Answer:
$ \boldsymbol{5} $

#### Absolute extrema:
# Brief Explanations:
As $ x \to \pm\infty $, $ y \to +\infty $, so there is no absolute maximum (the function goes to $ +\infty $). For the absolute minimum, since the function is continuous (polynomial) and as $ x \to \pm\infty $, $ y \to +\infty $, there must be an absolute minimum (the lowest point on the graph). So 1 absolute minimum, 0 absolute maximum.
# Answer:
$ \boldsymbol{1 \text{ (absolute minimum)}, 0 \text{ (absolute maximum)}} $ (or, if considering the question asks "how many absolute extrema", and since there's one absolute minimum and no absolute maximum, the number of absolute extrema is 1)

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Question

Explanation:

Part 1: Which function matches the graph?

Answer:

Part 2: Identify the characteristics (for the correct function \( f(x) = 2x^6 - 13x^5 + 26x^4 - 7x^3 - 28x^2 + 20x \))

Degree: