Problem 22 - Algebra and Trigonometry

Question

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. $$ f(x)=6 x^{2}+12 x-5 $$

Step-by-Step Solution

Answer

The vertex is (-1, -11), x-intercepts are approximately -0.31 and -2.69, and the y-intercept is (0, -5).

Step 1: Write the function in standard form

The standard form of a quadratic function is \( ax^2 + bx + c \). The given function \( f(x) = 6x^2 + 12x - 5 \) is already in standard form with \( a = 6 \), \( b = 12 \), and \( c = -5 \).

Step 2: Identify the coefficients

From the standard form \( f(x) = ax^2 + bx + c \), we identify \( a = 6 \), \( b = 12 \), and \( c = -5 \). We'll use these to find the vertex and intercepts.

Step 3: Find the vertex using the vertex formula

The vertex \((h, k)\) of a parabola can be found using the formula \( h = -\frac{b}{2a} \). Substituting \( a = 6 \) and \( b = 12 \), we get \( h = -\frac{12}{12} = -1 \). To find \( k \), substitute \( x = -1 \) back into \( f(x) \): \[ k = f(-1) = 6(-1)^2 + 12(-1) - 5 = 6 - 12 - 5 = -11 \]. Therefore, the vertex is \((-1, -11)\).

Step 4: Find the x-intercepts (roots)

Set \( f(x) = 0 \) to find the x-intercepts: \[ 6x^2 + 12x - 5 = 0 \]. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting \( a = 6 \), \( b = 12 \), and \( c = -5 \), we get: \[ x = \frac{-12 \pm \sqrt{12^2 - 4(6)(-5)}}{2(6)} = \frac{-12 \pm \sqrt{144 + 120}}{12} = \frac{-12 \pm \sqrt{264}}{12} \].Simplifying, the roots are approximately \( -0.31 \) and \(-2.69\).

Step 5: Find the y-intercept

The y-intercept is the value of \( f(x) \) when \( x = 0 \). Substitute \( x = 0 \) into the function: \[ f(0) = 6(0)^2 + 12(0) - 5 = -5 \].Thus, the y-intercept is \( (0, -5) \).

Step 6: Sketch the graph

Using the vertex \((-1, -11)\), the x-intercepts (approximately \(-0.31\) and \(-2.69\)), and the y-intercept \((0, -5)\), plot these points on a graph. Since the parabola opens upwards (as \( a = 6 > 0 \)), sketch the curve through these points with the axis of symmetry at \( x = -1 \).

Key Concepts

Standard Form

The standard form of a quadratic function is expressed as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. These constants dictate different characteristics of the parabola, such as its width, direction, and position. For our given function, \( f(x) = 6x^2 + 12x - 5 \), we can easily see that it is already in standard form because it fits the pattern.

• \( a = 6 \)
• \( b = 12 \)
• \( c = -5 \)

These values will not only help us in graphing the parabola but also in finding other important features like the vertex and intercepts. The standard form is a crucial starting point for analyzing quadratic functions because it gives a clear path to these key components.

Vertex

The vertex of a quadratic function is a crucial point that represents either the maximum or minimum value of the function, depending on the direction the parabola opens. For a function in standard form \( ax^2 + bx + c \), the \( x \)-coordinate of the vertex is calculated using the formula \( x = -\frac{b}{2a} \).
In our example, substituting \( a = 6 \) and \( b = 12 \), we find \( x = -\frac{12}{12} = -1 \). The \( y \)-coordinate, often denoted as \( k \), is found by substituting \( x = -1 \) back into the function: \( f(-1) = 6(-1)^2 + 12(-1) - 5 = -11 \).
Therefore, the vertex of this quadratic function is \((-1, -11)\), showing it is the lowest point because the parabola opens upwards (since \( a > 0 \)). It's the turning point of the function and gives symmetry to the parabola.

X-intercepts

The x-intercepts of a quadratic function are where the graph crosses the x-axis, and they can be found by setting the function \( f(x) = ax^2 + bx + c \) equal to zero and solving for \( x \). This can be done using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our function, substituting the values \( a = 6 \), \( b = 12 \), and \( c = -5 \) gives: \[ x = \frac{-12 \pm \sqrt{144 + 120}}{12} = \frac{-12 \pm \sqrt{264}}{12} \] By simplifying, the approximate intercepts are about \( x \approx -0.31 \) and \( x \approx -2.69 \).
These roots are also known as the solutions to the equation and give important information about the function's behavior.

Y-intercepts

The y-intercept of a quadratic function indicates where the graph crosses the y-axis. To find this, we evaluate the function \( f(x) = ax^2 + bx + c \) at \( x = 0 \).
For our function \( f(x) = 6x^2 + 12x - 5 \), the calculation is straightforward: \[ f(0) = 6(0)^2 + 12(0) - 5 = -5 \] This means the y-intercept is at the point \((0, -5)\). The y-intercept provides a starting point for graphing because it indicates the initial value of the function when no other variable is considered.
Keep in mind that unless \( b = 0 \), the y-intercept is unique since the quadratic term and the linear term become zero when \( x = 0 \). This makes the y-intercept the constant term \( c \).