Chapter 4: Problem 21
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. $$ f(x)=-4 x^{2}-16 x+3 $$
Short Answer
Expert verified
Vertex: (2, -45), x-intercepts: (-0.19, 0) and (-3.81, 0), y-intercept: (0, 3); parabola opens downwards.
Step by step solution
01
Write Quadratic in Standard Form
The given quadratic function is already in standard form, which is \( f(x) = ax^2 + bx + c \). Here, \( a = -4 \), \( b = -16 \), and \( c = 3 \).
02
Find the Vertex
For a quadratic function \( f(x) = ax^2 + bx + c \), the vertex \((h, k)\) can be found using the formulas \( h = -\frac{b}{2a} \) and \( k = f(h) \).First, find \( h \):\[ h = -\frac{-16}{2(-4)} = 2 \\]Then, find \( k \):\[ k = f(2) = -4(2)^2 - 16(2) + 3 = -16 - 32 + 3 = -45 \\]Thus, the vertex is \((2, -45)\).
03
Calculate the x-intercepts
The x-intercepts occur where \( f(x) = 0 \). Solve the equation \(-4x^2 - 16x + 3 = 0\) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).\[ b^2 - 4ac = (-16)^2 - 4(-4)(3) = 256 + 48 = 304 \ x = \frac{-(-16) \pm \sqrt{304}}{2(-4)} = \frac{16 \pm \sqrt{304}}{-8} \\]Calculate the values:- First root: \( x_1 = \frac{16 + \sqrt{304}}{-8} \)- Second root: \( x_2 = \frac{16 - \sqrt{304}}{-8} \)These individual calculations yield approximately \( x_1 \approx -0.19 \) and \( x_2 \approx -3.81 \).
04
Determine the y-intercept
The y-intercept occurs where \( x = 0 \). Calculate \( f(0) \):\[ f(0) = -4(0)^2 - 16(0) + 3 = 3 \\]Thus, the y-intercept is \( (0, 3) \).
05
Sketch the Graph
To sketch the graph:1. Plot the vertex at \((2, -45)\).2. Mark the x-intercepts at approximately \((-0.19, 0)\) and \((-3.81, 0)\).3. Plot the y-intercept at \((0, 3)\).4. Note that the parabola opens downwards since \( a = -4 \), indicating it's a downward facing concave parabola.5. Draw a smooth curve passing through these points.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex Form
The vertex form of a quadratic function makes it easier to identify the vertex of the parabola quickly. It is given by the expression \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex of the parabola. For the quadratic function \( f(x) = -4x^2 - 16x + 3 \), the vertex can be calculated using the formulas for finding \( h \) and \( k \).
- First, find the value of \( h \) by using the formula \( h = -\frac{b}{2a} \). In this case, it simplifies to \( h = -\frac{-16}{2(-4)} = 2 \).
- Next, substitute \( h \) back into the function to find \( k \). So, \( k = f(2) = -45 \).
The vertex of this parabola is \((2, -45)\), and knowing this helps us understand where the lowest or highest point of the graph is located. This is particularly useful when graphing parabolas or understanding how the graph shifts.
- First, find the value of \( h \) by using the formula \( h = -\frac{b}{2a} \). In this case, it simplifies to \( h = -\frac{-16}{2(-4)} = 2 \).
- Next, substitute \( h \) back into the function to find \( k \). So, \( k = f(2) = -45 \).
The vertex of this parabola is \((2, -45)\), and knowing this helps us understand where the lowest or highest point of the graph is located. This is particularly useful when graphing parabolas or understanding how the graph shifts.
X-intercepts
X-intercepts are the points where the graph of the quadratic function crosses the x-axis. These occur where \( f(x) = 0 \). Finding these points can help in sketching the graph of the function. Solving the equation \(-4x^2 - 16x + 3 = 0\) gives the x-intercepts.
- The method of solving for x-intercepts often involves the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
- For our quadratic: calculate the part under the square root, \( b^2 - 4ac \). This gives \( 304 \) in the expression.
- Solve to find the roots, \( x_1 \approx -0.19 \) and \( x_2 \approx -3.81 \).
Y-intercepts
The y-intercept of a function is the point where the graph crosses the y-axis. This can provide a reference point when graphing or analyzing the function. For any quadratic function of the form \( f(x) = ax^2 + bx + c \), the y-intercept occurs where \( x = 0 \). This simplifies the determination of the y-intercept.
For our function \( f(x) = -4x^2 - 16x + 3 \):
- Substitute \( x = 0 \) into the function to find \( f(0) = 3 \).
Thus, the y-intercept is \((0, 3)\). This point is essential as it shows where the parabola begins its journey on the y-axis, providing a starting point while sketching the graph.
For our function \( f(x) = -4x^2 - 16x + 3 \):
- Substitute \( x = 0 \) into the function to find \( f(0) = 3 \).
Thus, the y-intercept is \((0, 3)\). This point is essential as it shows where the parabola begins its journey on the y-axis, providing a starting point while sketching the graph.
Graphing Parabolas
Graphing parabolas involves using key points such as the vertex, x-intercepts, and y-intercept to draw a smooth curve that represents the function. Understanding the direction in which the parabola opens is crucial as well.
- First, plot the vertex \((2, -45)\), which serves as the "tip" of the parabola.
- Next, mark the x-intercepts approximately at \((-0.19, 0)\) and \((-3.81, 0)\).
- Find and plot the y-intercept at \((0, 3)\).