Graphing Polynomial Functions

Given, f(x) = -2x³ + 6x² – 4x
Finding Degree and Leading Coefficient
Since the degree is 3 (odd), and the leading coefficient is -2 (positive), then
As x → ∞, f(x) → -∞
As x → -∞, f(x) → ∞
Finding Zeros
Now, f(x) = -2x³ + 6x² – 4x
⇒ f(x) = -2x(x² – 3x + 2)
⇒ f(x) = -2x(x² – 2x – x + 2)
⇒ f(x) = -2x[x(x – 2) – 1(x – 2)]
⇒ f(x) = -2x(x – 2)(x – 1)
Now, putting f(x) = 0 ⇒ -2x(x – 2)(x – 1) = 0 gives
x = 0, 1, and 2
Thus, the x-intercepts are at (0, 0), (1, 0), and (2, 0)
Finding the Y-Intercept 
By substituting x = 0, we get
f(0) = -2(0)³ + 6(0)² – 4(0)
⇒ f(0) = 0
Thus, the y-intercept is at (0, 0)
Determining Turning Points
The first derivative is 
f’(x) = -6x² + 12x – 4
-6x² + 12x – 4 = 0
⇒ 3x² – 6x + 2 = 0
⇒ x = ${\dfrac{-\left( -6\right) \pm \sqrt{\left( -6\right) ^{2}-4\times 3\times 2}}{2\times 3}}$
⇒ x = ${\dfrac{6\pm \sqrt{36-24}}{6}}$
⇒ x = ${\dfrac{6\pm \sqrt{12}}{6}}$
⇒ x = ${\dfrac{6\pm 2\sqrt{3}}{6}}$
⇒ x = ${1\pm \dfrac{\sqrt{3}}{3}}$
Since ${1+\dfrac{\sqrt{3}}{3}}$ ≈ 1.58 and ${1-\dfrac{\sqrt{3}}{3}}$ ≈ 0.42
At x = 1.58, f(x) = 0.77
At x = 0.42, f(x) = -0.77
Thus, the critical points are at (1.58, 0.77) and (0.42, -0.77)
Identifying the Concavity and Inflection Point 
The second derivative is
f”(x) = -12x + 12
-12x + 12 = 0
⇒ 12x = 12
⇒ x = 1
Thus, the inflection point is (1, 0)
After plotting these points, we get the required graph.

Given, f(x) = -x² + 4x – 3
Finding the Degree and Leading Coefficient
Here, f(x) is a quadratic function in the standard form:
f(x) = ax² + bx + c
a = -1, b = 4, and c = -3
Thus, the degree is 2 (even), and the leading coefficient is -1 (negative).
End behavior:
As x → ∞, f(x) → -∞
As x → -∞, f(x) → -∞ 
Finding the Roots (Zeros)
Setting f(x) = 0, we get
-x² + 4x – 3 = 0
⇒ x² – 4x + 3 = 0
⇒ x² – 3x – x + 3 = 0
⇒ x(x – 3) – 1 (x – 3) = 0
⇒ (x – 3)(x – 1) = 0
⇒ x = 3 and 1
Thus, the x-intercepts are at (1, 0) and (3, 0)
Finding the Y-Intercept
When x = 0,
f(0) = -(0)² + 4(0) – 3 
⇒ f(0) = -3
Thus, the y-intercept is at (0, -3)
Finding the Vertex
The formula for the x-coordinate of the vertex is:
${x=-\dfrac{b}{2a}}$
Substituting a = -1 and b = 4, 
${x=-\dfrac{4}{2\left( -1\right) }}$
⇒ x = 2
At x = 2, f(x) = -(2)² + 4(2) – 3 = 1
Thus, the vertex is at (2, 1)
After plotting these points, we get the required graph.

Here, we follow the same steps, such as graphing cubic polynomial functions.
Given, f(x) = x⁴ – 4x² 
Finding the Degree and the Leading Coefficient
Since the degree is 4 (even), and the leading coefficient is 1 (positive), then
As x → ∞, f(x) → ∞
As x → -∞, f(x) → ∞
Finding Zeros
Now, f(x) = x⁴ – 4x² 
⇒ f(x) = x²(x² – 4)
⇒ f(x) = x²(x + 2)(x – 2)
Now, putting f(x) = 0 ⇒ x²(x + 2)(x – 2) = 0 gives
x = 0, -2, and 2
Thus, the x-intercepts are at (-2, 0), (0, 0), and (2, 0)
Finding the Y-Intercept
By substituting x = 0, we get
f(0) = (0)⁴ – 4(0)² 
⇒ f(0) = 0
Thus, the y-intercept is at (0, 0)
Determining Turning Points
Now, the first derivative is f’(x) = 4x³ – 8x
4x³ – 8x = 0
⇒ 4x(x² – 2) = 0
⇒ x = 0, ${\sqrt{2}}$, and ${-\sqrt{2}}$
At x = 0, f(x) = 0
At x = ${\sqrt{2}}$, f(x) = -4
At x = ${-\sqrt{2}}$, f(x) = -4
Since ${\sqrt{2}}$ ≈ 1.41 and ${-\sqrt{2}}$ ≈ -1.41
Thus, the critical points are (0, 0), (1.41, -4), and (-1.41, -4)
Identifying the Concavity and Inflection Point
The second derivative is f”(x) = 12x² – 8
12x² – 8 = 0
⇒ 12x² = 8
⇒ x² = ${\dfrac{2}{3}}$
⇒ x = ${\pm \sqrt{\dfrac{2}{3}}}$ ≈ ±0.82
At x = ±0.82, f(x) = -2.22
Thus, inflection points are at (0.82, -2.22) and (-0.82, -2.22)
After plotting all these points, we get the required graph.

Here, from the graph provided:
Shape of the graph: The graph is a parabola, which suggests it is a degree-2 polynomial function, i.e., a quadratic function.
Vertex: The highest point (vertex) of the parabola is at (0, 4)
Roots (x-intercepts): The graph crosses the x-axis at (-2, 0) and (2, 0)
As we know, the vertex form of a quadratic function is: y = a(x – h)² + k …..(i)
Here, (h, k) is the vertex
Substituting the vertex (0,4) in the formula (i), we get
y = a(x – 0)² + 4
⇒ y = ax² + 4 …..(ii)
Now, substituting one of the roots (2, 0) in the equation (ii), we get
0 = a(2)² + 4
⇒ 0 = 4a + 4
⇒ 4a = -4
⇒ a = -1
Substituting the value a = -1 in the equation (ii), we get
y = -x² + 4
Thus, the equation of the polynomial function is y = -x² + 4

Here, 
The graph is a parabola that opens upward, indicating the leading coefficient (i.e., a) is positive.
The vertex is at (h, k) = (-3, -1)
The x-intercepts are (-4, 0) and (-2, 0)
The y-intercept is at (0, 10)
As we know, the parabola represents a quadratic polynomial function, and its standard form is:
y = a(x – h)² + k …..(i)
Here, substituting the vertex in the equation (i), we get
y = a[x – (-3)]² + (-1)
⇒ y = a(x + 3)² – 1 …..(ii)
Now, substituting one of the roots (-2, 0) in the equation (ii), we get
0 = a(-2 + 3)² – 1
⇒ a – 1 = 0
⇒ a = 1
Thus, the equation (i) becomes
y = 1(x + 3)² – 1
⇒ y = x² + 6x + 9 – 1
⇒ y = x² + 6x + 8, which is option c.
Thus, option c) represents the given graph.

Since even-degree polynomial graphs:
1. Have symmetrical ends (either both ends point up or both point down).
2. It can have zero, two, or more turning points.
3. The number of x-intercepts varies but is never odd.

Since odd-degree polynomial graphs:
1. Have ends pointing in opposite directions (one end up, the other down).
2. They pass through the origin if the constant term is zero.

Thus, 
Option a) and c) represent even-degree polynomial functions.
Option b) and d) represent odd-degree polynomial functions.