By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
By the end of this topic, students will be able to: - Identify and describe the characteristics of quadratic, cubic, and reciprocal graphs. - Sketch and interpret graphs of quadratic, cubic, and reciprocal functions. - Use the properties of these graphs to solve problems and answer questions. - Apply algebraic techniques to find equations of quadratic, cubic, and reciprocal graphs. - Analyze and compare the features of different types of graphs.
Quadratic graphs are U-shaped and can be written in the form $y = ax^2 + bx + c$. The vertex of a quadratic graph is the point where the graph changes direction. The x-coordinate of the vertex can be found using the formula $x = -\frac{b}{2a}$.
Cubic graphs are more complex and can have multiple turning points. They can be written in the form $y = ax^3 + bx^2 + cx + d$. Cubic graphs can have one, two, or three turning points.
Reciprocal graphs have a vertical asymptote and can be written in the form $y = \frac{1}{x}$. Reciprocal graphs can be transformed using the rules of function notation.
Sketch the graph of $y = x^2 - 4x + 3$.
To sketch this graph, we need to find the x-coordinate of the vertex using the formula $x = -\frac{b}{2a}$. In this case, $a = 1$ and $b = -4$, so $x = -\frac{-4}{2(1)} = 2$. We can then substitute this value into the equation to find the y-coordinate: $y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$. The vertex of the graph is therefore $(2, -1)$.
The graph will be a U-shape that opens upwards, since $a$ is positive. To find the x-intercepts, we need to solve the equation $x^2 - 4x + 3 = 0$. We can factor the equation as $(x - 3)(x - 1) = 0$, which gives us the x-intercepts $x = 3$ and $x = 1$.
The graph of $y = x^3 - 6x^2 + 11x - 6$ has a turning point at $(2, 0)$. Find the equation of the graph.
To find the equation of the graph, we need to use the fact that the turning point is given by the formula $x = -\frac{b}{2a}$. In this case, $a = 1$ and $b = -6$, so $x = -\frac{-6}{2(1)} = 3$. We can then substitute this value into the equation to find the y-coordinate: $y = (3)^3 - 6(3)^2 + 11(3) - 6 = 27 - 54 + 33 - 6 = 0$. The turning point is therefore $(3, 0)$.
We can then use the fact that the graph has a turning point at $(3, 0)$ to write the equation in the form $y = a(x - 3)^2 + b$. We can then substitute the point $(2, 0)$ into the equation to find the values of $a$ and $b$. This gives us the equation $y = a(x - 3)^2 + b$, which we can then expand to find the equation of the graph.
Sketch the graph of $y = \frac{1}{x + 2}$.
To sketch this graph, we need to find the vertical asymptote, which occurs when the denominator is equal to zero. In this case, the denominator is $x + 2$, so the vertical asymptote occurs when $x + 2 = 0$, which gives us $x = -2$.
The graph will have a vertical asymptote at $x = -2$ and will approach the x-axis as $x$ approaches infinity. To find the x-intercept, we need to solve the equation $\frac{1}{x + 2} = 0$. This equation has no solution, since the denominator is never equal to zero.
What is the x-coordinate of the vertex of the graph of $y = x^2 - 4x + 3$? A) 1 B) 2 C) 3 D) 4
Correct answer: B) 2 Why the distractors fail: A) 1 is the x-intercept, not the vertex. C) 3 is the x-intercept, not the vertex. D) 4 is not a valid option.
What is the equation of the graph of $y = x^3 - 6x^2 + 11x - 6$ if the turning point is at $(2, 0)$? A) $y = x^3 - 6x^2 + 11x - 6$ B) $y = (x - 2)^3 - 6(x - 2)^2 + 11(x - 2) - 6$ C) $y = (x - 3)^3 - 6(x - 3)^2 + 11(x - 3) - 6$ D) $y = (x - 2)^2 - 6(x - 2) + 11$
Correct answer: C) $y = (x - 3)^3 - 6(x - 3)^2 + 11(x - 3) - 6$ Why the distractors fail: A) is the original equation, not the equation of the graph with the turning point at $(2, 0)$. B) is a possible equation, but not the correct one. D) is a quadratic equation, not a cubic equation.
What is the vertical asymptote of the graph of $y = \frac{1}{x + 2}$? A) $x = -1$ B) $x = -2$ C) $x = 1$ D) $x = 2$
Correct answer: B) $x = -2$ Why the distractors fail: A) is not a valid option. C) and D) are not the vertical asymptotes.
What is the x-intercept of the graph of $y = x^2 - 4x + 3$? A) $x = 1$ B) $x = 2$ C) $x = 3$ D) $x = 4$
Correct answer: C) $x = 3$ Why the distractors fail: A) is not the x-intercept. B) is the x-coordinate of the vertex, not the x-intercept. D) is not a valid option.
What is the type of graph of $y = \frac{1}{x}$? A) Quadratic B) Cubic C) Reciprocal D) Linear
Correct answer: C) Reciprocal Why the distractors fail: A) and B) are not the correct types of graphs. D) is not a valid option.
Sketch the graph of $y = x^2 - 4x + 3$ and identify its characteristics.
Find the equation of the graph of $y = x^3 - 6x^2 + 11x - 6$ if the turning point is at $(2, 0)$.
Sketch the graph of $y = \frac{1}{x + 2}$ and identify its characteristics.
Find the x-intercepts of the graph of $y = x^2 - 4x + 3$.
Find the equation of the graph of $y = x^2 - 4x + 3$ if the vertex is at $(2, -1)$.
Note: The short-answer questions are designed to assess the students' ability to apply the concepts and techniques learned in the topic.
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