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Chapter 4: Problem 10

A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its x- and y-intercept(s). (c) Sketch its graph. $$ f(x)=x^{2}+8 x $$

Short Answer

Expert verified
The vertex is \((-4, -16)\), x-intercepts are \((0, 0)\) and \((-8, 0)\), and y-intercept is \((0, 0)\). The graph is a parabola opening upwards passing through these points.

Step by step solution

01

Express in Standard Form

The standard form of a quadratic function is given by the equation \( f(x) = ax^2 + bx + c \). For the function \( f(x) = x^2 + 8x \), we can identify the coefficients: \( a = 1 \), \( b = 8 \), and \( c = 0 \). Therefore, it is already in standard form: \( f(x) = x^2 + 8x \).
02

Find the Vertex

The vertex of a quadratic function \( f(x) = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: \( x = -\frac{b}{2a} \). For the function \( f(x) = x^2 + 8x \), substitute \( a = 1 \) and \( b = 8 \) into the formula: \( x = -\frac{8}{2(1)} = -4 \). Substitute \( x = -4 \) back into the function to find the y-coordinate: \( f(-4) = (-4)^2 + 8(-4) = 16 - 32 = -16 \). Thus, the vertex is \((-4, -16)\).
03

Find the X-Intercepts

The x-intercepts can be found by setting \( f(x) = 0 \) and solving for \( x \). Set \( x^2 + 8x = 0 \), and factor out \( x \): \( x(x + 8) = 0 \). This gives the solutions \( x = 0 \) and \( x = -8 \), so the x-intercepts are \((0, 0)\) and \((-8, 0)\).
04

Find the Y-Intercept

The y-intercept is the value of \( f(x) \) when \( x = 0 \). Substitute \( x = 0 \) into the function: \( f(0) = 0^2 + 8(0) = 0 \). Thus, the y-intercept is \((0, 0)\).
05

Sketch the Graph

To sketch the graph of the function \( f(x) = x^2 + 8x \), plot the vertex \((-4, -16)\), the x-intercepts \((0, 0)\) and \((-8, 0)\), and the y-intercept \((0, 0)\). Draw a parabola opening upwards since the coefficient of \( x^2 \) is positive. The axis of symmetry is along \( x = -4 \), passing through the vertex.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertex of a Parabola
The vertex of a parabola is a crucial point on its graph. It represents the tip or the lowest/highest point, depending on whether the parabola opens upwards or downwards. The vertex can be found using the formula for its x-coordinate:
  • For a quadratic function of the form \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \).
Let's consider the quadratic function \( f(x) = x^2 + 8x \):
  • Here, \( a = 1 \) and \( b = 8 \).
  • Plug these values into the formula: \( x = -\frac{8}{2(1)} = -4 \).
Once you have the x-coordinate, insert it back into the function to get the y-coordinate:
  • Calculating, \( f(-4) = (-4)^2 + 8(-4) = 16 - 32 = -16 \).
So the vertex is \((-4, -16)\). This vertex tells you the parabola's minimum point, and the fact that \( a \) is positive indicates that the parabola opens upwards.
X-intercepts of a Quadratic Function
X-intercepts are the points where the graph of the function crosses the x-axis. For a quadratic function, these are the solutions to the equation \( f(x) = 0 \). Let's find the x-intercepts of the quadratic function \( f(x) = x^2 + 8x \):
  • Set the equation to zero: \( x^2 + 8x = 0 \).
  • Factor the equation: \( x(x + 8) = 0 \).
  • This gives the x-intercepts: \( x = 0 \) and \( x = -8 \).
These solutions mean that the points \((0, 0)\) and \((-8, 0)\) are where the graph cuts through the x-axis.
The x-intercepts are important for understanding the roots of the quadratic equation and provide essential points for sketching its graph.
Y-intercepts of a Function
The y-intercept is where the graph crosses the y-axis. This occurs when all x values are zero, meaning you substitute \( x = 0 \) into the function to find the intercept. For \( f(x) = x^2 + 8x \), the calculation is straightforward:
  • Substitute \( x = 0 \) into the function: \( f(0) = 0^2 + 8(0) = 0 \).
Therefore, the y-intercept is \((0, 0)\).
It's a special point that not only tells you where the graph meets the y-axis but is also one of the x-intercepts in this particular case. Graphically, it provides a starting point for sketching the quadratic function.
Standard Form of a Quadratic
The standard form of a quadratic function is expressed as \( f(x) = ax^2 + bx + c \). This is one of the canonical forms of quadratics, helping to easily identify key characteristics of the function:
  • The coefficient \( a \) tells you which direction the parabola opens (upwards if \( a > 0 \), downwards if \( a < 0 \)).
  • The term \( b \) contributes to the parabola's horizontal placement.
  • The constant \( c \) can be seen as the y-intercept when \( x = 0 \) since \( f(0) = c \).
For the quadratic function \( f(x) = x^2 + 8x \), the coefficients are:
  • \( a = 1 \)
  • \( b = 8 \)
  • \( c = 0 \)
This function is already in standard form. Recognizing the standard form helps in simplifying the process of finding other features like the vertex, axis of symmetry, and intercepts, and also aids in graphing the function.

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Most popular questions from this chapter

Transformations of \(y=1 / x^{2}\) In Example 2 we saw that some simple rational functions can be graphed by shifting, stretching, or reflecting the graph of \(y=1 / x\) . In this exercise we consider rational functions that can be graphed by transforming the graph of \(y=1 / x^{2},\) shown on the following page. (a) Graph the function $$ r(x)=\frac{1}{(x-2)^{2}} $$ by transforming the graph of \(y=1 / x^{2}\) . (b) Use long division and factoring to show that the function $$ s(x)=\frac{2 x^{2}+4 x+5}{x^{2}+2 x+1} $$ can be written as $$ s(x)=2+\frac{3}{(x+1)^{2}} $$ Then graph \(s\) by transforming the graph of \(y=1 / x^{2} .\) (c) One of the following functions can be graphed by trans- forming the graph of \(y=1 / x^{2} ;\) the other cannot. Use transformations to graph the one that can be, and explain why this method doesn't work for the other one. $$ p(x)=\frac{2-3 x^{2}}{x^{2}-4 x+4} \quad q(x)=\frac{12 x-3 x^{2}}{x^{2}-4 x+4} $$

Show that the given values for \(a\) and \(b\) are lower and upper bounds for the real zeros of the polynomial. $$ P(x)=x^{4}-2 x^{3}-9 x^{2}+2 x+8 ; \quad a=-3, b=5 $$

Find all rational zeros of the polynomial, and write the polynomial in factored form. $$ P(x)=6 x^{4}-7 x^{3}-12 x^{2}+3 x+2 $$

Focusing Distance For a camera with a lens of fixed fo- cal length \(F\) to focus on an object located a distance \(x\) from the lens, the film must be placed a distance \(y\) behind the lens, where \(F, x,\) and \(y\) are related by $$ \frac{1}{x}+\frac{1}{y}=\frac{1}{F} $$ (See the figure.) Suppose the camera has a \(55-\mathrm{mm}\) lens \((F=55) .\) (a) Express \(y\) as a function of \(x\) and graph the function. (b) What happens to the focusing distance \(y\) as the object moves far away from the lens? (c) What happens to the focusing distance \(y\) as the object moves close to the lens?

Find all rational zeros of the polynomial, and then find the irrational zeros, if any. Whenever appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartes' Zule of Signs, the Uuadratic formula, or other factoring techniques. $$ P(x)=6 x^{4}-7 x^{3}-8 x^{2}+5 x $$
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