Problem 19 - Algebra A Combined Function

Question

Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and graph the function. $$ f(x)=4 x^{2}+4 x-3 $$

Step-by-Step Solution

Answer

The vertex is \((-\frac{1}{2}, -4)\), the parabola opens upward, the y-intercept is \((0, -3)\), and the x-intercepts are \((\frac{1}{2}, 0)\) and \((-\frac{3}{2}, 0)\).

Step 1: Identify the Standard Form

The given quadratic function is \(f(x) = 4x^2 + 4x - 3\). This is in the standard form \(ax^2 + bx + c\), where \(a = 4\), \(b = 4\), and \(c = -3\).

Step 2: Find the Vertex Using Vertex Formula

The x-coordinate of the vertex is given by the formula \(x = -\frac{b}{2a}\). Substitute \(b = 4\) and \(a = 4\) into the formula:\[x = -\frac{4}{2 \times 4} = -\frac{1}{2}\]To find the y-coordinate, substitute \(x = -\frac{1}{2}\) back into the function:\[f\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{2}\right)^2 + 4\left(-\frac{1}{2}\right) - 3\]\[= 4\times\frac{1}{4} - 2 - 3\]\[= 1 - 2 - 3 = -4\]The vertex is \((-\frac{1}{2}, -4)\).

Step 3: Determine the Direction the Parabola Opens

The direction in which the parabola opens is determined by the coefficient \(a\). Since \(a = 4\), which is greater than 0, the parabola opens upward.

Step 4: Find the Intercepts

To find the y-intercept, substitute \(x = 0\) into the function: \[f(0) = 4 \times 0^2 + 4 \times 0 - 3 = -3\] So, the y-intercept is \((0, -3)\).To find the x-intercepts, set the function equal to zero and solve for \(x\):\[4x^2 + 4x - 3 = 0\]Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(b = 4\), \(a = 4\), and \(c = -3\):\[x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4}\]\[x = \frac{-4 \pm \sqrt{16 + 48}}{8}\]\[x = \frac{-4 \pm \sqrt{64}}{8}\]\[x = \frac{-4 \pm 8}{8}\]The solutions are \(x = \frac{4}{8} = \frac{1}{2}\) and \(x = \frac{-12}{8} = -\frac{3}{2}\), so the x-intercepts are \((\frac{1}{2}, 0)\) and \((-\frac{3}{2}, 0)\).

Step 5: Plot the Graph

To graph the function, plot the vertex \((-\frac{1}{2}, -4)\), the y-intercept \((0, -3)\), and the x-intercepts \((\frac{1}{2}, 0)\) and \((-\frac{3}{2}, 0)\). Since the parabola opens upward, sketch the curve passing through these points.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial point that helps you understand the shape and position of the graph. In the standard quadratic function form, which is expressed as \( ax^2 + bx + c \), the vertex can be determined using the vertex formula for the x-coordinate: \( x = -\frac{b}{2a} \). This formula highlights the symmetry of the parabola.

For the quadratic function \( f(x) = 4x^2 + 4x - 3 \), we have \( a = 4 \), \( b = 4 \), and \( c = -3 \). Using the vertex formula, we find the x-coordinate of the vertex to be \( x = -\frac{1}{2} \).

Substituting \( x = -\frac{1}{2} \) back into the function gives us the y-coordinate:

• Calculate \( 4\left(-\frac{1}{2}\right)^2 = 1 \)
• \( 4\left(-\frac{1}{2}\right) = -2 \)
• Combine everything to find \( 1 - 2 - 3 = -4 \)

Thus, the vertex of the parabola is \(-\frac{1}{2}, -4\). This point is often the minimum or maximum point of the parabola, where the direction changes.

Intercepts of Quadratic Function

Intercepts in a quadratic function give you key information on where the graph crosses the axes. This includes both the x-intercepts and the y-intercept.

To find the **y-intercept**, you set \( x = 0 \). For our function \( f(x) = 4x^2 + 4x - 3 \), substituting \( x = 0 \) gives us \( f(0) = -3 \). Therefore, the y-intercept is at \( (0, -3) \), meaning the graph crosses the vertical axis at this point.

For the **x-intercepts**, where the graph crosses the horizontal axis, set the entire function equal to zero: \( 4x^2 + 4x - 3 = 0 \). Solving this using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 4 \), \( b = 4 \), and \( c = -3 \), yields the solutions:

• \( x = \frac{1}{2} \)
• \( x = -\frac{3}{2} \)

These x-intercepts give us the points \( (\frac{1}{2}, 0) \) and \( (-\frac{3}{2}, 0) \). Recognizing these intercepts helps when sketching the graph, as they are the points where the function touches or crosses the x-axis.

Direction of Parabola Opening

The direction in which a parabola opens is easy to determine and directly relates to the coefficient \( a \) in the quadratic function \( ax^2 + bx + c \). This value tells you whether the parabola goes upwards or downwards.

If \( a \) is positive, the parabola opens **upward**, resembling a 'U' shape. For our function \( f(x) = 4x^2 + 4x - 3 \), with \( a = 4 \), the parabola opens upwards because 4 is greater than zero. Conversely, if \( a \) were negative, the parabola would open **downward**, like an upside-down 'U'.

This aspect affects the vertex as well:

• If the parabola opens upward, the vertex represents the minimum point on the graph.
• If it opens downward, the vertex is the maximum point.

Understanding the direction of opening is essential when plotting or predicting the behavior of the quadratic graph, especially in applications like physics and engineering.