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Chapter 3: Problem 23

Graphing Quadratic Functions A quadratic function \(f\) is given. (a) Express \(f\) in standard form. (b) Find the vertex and \(x\) and \(y\) -intercepts of \(f .\) (c) Sketch a graph of \(f .\) (d) Find the domain and range of \(f\). $$f(x)=-4 x^{2}-12 x+1$$

Short Answer

Expert verified
The quadratic function in standard form is \(f(x) = -4(x+1.5)^2 + 10\). The vertex is \((-1.5, 10)\), x-intercepts are approximately \(0.183\) and \(-1.366\), y-intercept is \(1\), domain is all real numbers, and range is \((-\infty, 10]\).

Step by step solution

01

Converting to Standard Form

The standard form of a quadratic function is \( f(x) = a(x-h)^2 + k \). To convert \( f(x)=-4x^2-12x+1 \) to standard form, we need to complete the square. First, factor out \(-4\) from the \(x\)-terms: \( f(x) = -4(x^2 + 3x) + 1 \). Then, complete the square inside the parentheses: \( f(x) = -4(x^2 + 3x + \frac{9}{4} - \frac{9}{4}) + 1 = -4((x + \frac{3}{2})^2 - \frac{9}{4}) + 1 \). Simplify to obtain \( f(x) = -4(x + \frac{3}{2})^2 + 10 \). Thus, the standard form is \( f(x) = -4(x + 1.5)^2 + 10 \).
02

Finding the Vertex

The vertex form of a quadratic function \( f(x) = a(x-h)^2 + k \) makes it easy to identify the vertex, which is \((h, k)\). From \( f(x) = -4(x+\frac{3}{2})^2 + 10 \), \( h = -\frac{3}{2} \) and \( k = 10 \). Therefore, the vertex is \((-\frac{3}{2}, 10)\).
03

Finding the X-intercepts

To find the \(x\)-intercepts, set \( f(x) = 0 \). Solve \(-4x^2 - 12x + 1 = 0\) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For \( a = -4, b = -12, c = 1 \), the discriminant \( b^2 - 4ac = 144 - 16 = 128 \). The solutions are \( x = \frac{12 \pm \sqrt{128}}{-8} \). Simplify to find \( x = 0.183 \) and \( x = -1.366 \). Thus, the \(x\)-intercepts are approximately \(0.183\) and \(-1.366\).
04

Finding the Y-intercept

To find the \(y\)-intercept, set \(x = 0\) in \( f(x) \). Substitute to get \( f(0) = -4(0)^2 - 12(0) + 1 = 1 \). The \(y\)-intercept is \((0, 1)\).
05

Sketching the Graph

Using the vertex \((-\frac{3}{2}, 10)\), x-intercepts \((0.183, 0)\) and \((-1.366, 0)\), and the y-intercept \((0, 1)\), sketch a downward opening parabola (since \(a = -4\)), ensuring that it passes through the intercepts and vertex.
06

Finding Domain and Range

The domain of any quadratic function is all real numbers \(( -\infty, +\infty)\). Since the parabola opens downward (\(a < 0\)) with the vertex at \(y=10\), the range is \(( -\infty, 10]\).

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Key Concepts

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

Standard Form of a Quadratic Function
A quadratic function can be expressed in standard form as \( f(x) = a(x-h)^2 + k \). This form is very useful because it easily reveals the graph's transformations, such as shifts and stretches. To convert a quadratic function like \( f(x)=-4x^2-12x+1 \) into this form, we use a method known as "completing the square."
  • First, factor out the \(a\) coefficient (which is -4 in this case) from the quadratic and linear terms.
  • Next, complete the square inside the parentheses. This involves adding and subtracting the right constant to turn the expression into a perfect square trinomial.
  • Simplify the expression to reshape it into the standard form.
For the given function, this process results in \( f(x) = -4(x + \frac{3}{2})^2 + 10 \). This form allows us to easily identify critical features of the quadratic graph like the vertex.
Vertex of a Quadratic Function
The vertex of a quadratic function is a key point that indicates the highest or lowest point of the graph, depending on whether it opens upward or downward. In the standard form \( f(x) = a(x-h)^2 + k \), the vertex is located at \((h, k)\).
For our example \( f(x) = -4(x + \frac{3}{2})^2 + 10 \), we can see:
  • \(h = -\frac{3}{2}\) and \(k = 10\).
Thus, the vertex is at \((-\frac{3}{2}, 10)\). Since the coefficient \(a\) is negative (-4), the parabola opens downward, and the vertex represents the maximum point on the graph.
X-intercepts and Y-intercepts
Intercepts are points where the graph crosses the axes. They are essential to graphing a quadratic function:
  • **X-intercepts**: Set \( f(x) = 0 \) and solve for \(x\). This means solving \(-4x^2 - 12x + 1 = 0\). By using the quadratic formula, the solutions are approximately \(x = 0.183\) and \(x = -1.366\).
  • **Y-intercept**: Substitute \(x = 0\) into the function. For \( f(x) = -4x^2 - 12x + 1 \), setting \(x = 0\) results in \( f(0) = 1 \). Thus, the \(y\)-intercept is \((0, 1)\).
These intercepts help in sketching the graph accurately. They provide points the parabola must pass through, creating a more precise image of its shape.
Domain and Range of Quadratic Functions
Quadratic functions have specific domain and range characteristics that define possible input and output values:
  • **Domain**: For any quadratic function, the domain is all real numbers, \(( -\infty, +\infty)\), because you can substitute any value of \(x\) into the equation.
  • **Range**: Depending on the direction the parabola opens, the range can differ. Since our given function \( f(x) = -4(x + \frac{3}{2})^2 + 10 \) has a negative \(a\), it opens downward. Thus, the range is limited from \( -\infty \) to the maximum point, which is the vertex at \( y = 10 \). Therefore, the range is \((-\infty, 10]\).
Understanding these concepts aids in constructing the graph and interpreting potential real-world applications of quadratic functions.

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