Problem 13 - 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. $$ f(x)=x^{2}+4 x-5 $$

Step-by-Step Solution

Answer

Vertex: (-2, -9); opens upward; intercepts: x=(-5,0),(1,0), y=(0,-5).

Step 1: Identify the coefficients

In the quadratic function \(f(x) = x^2 + 4x - 5\), the coefficients are identified as follows: \(a = 1\), \(b = 4\), and \(c = -5\).

Step 2: Determine the vertex using the vertex formula

The vertex \((h, k)\) of a parabola given by a quadratic function \(f(x) = ax^2 + bx + c\) can be found using the formula \(h = -\frac{b}{2a}\). Here, \(h = -\frac{4}{2 \times 1} = -2\). Substitute \(x = -2\) into the function to find \(k\): \(k = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9\). Thus, the vertex is \((-2, -9)\).

Step 3: Determine the direction of the parabola

A parabola opens upward if \(a > 0\) and downward if \(a < 0\). Since \(a = 1\) (which is positive), the graph of the function opens upward.

Step 4: Find the x-intercepts

To find the x-intercepts, set \(f(x) = 0\) and solve for \(x\): \(x^2 + 4x - 5 = 0\). Factoring the quadratic equation gives \((x + 5)(x - 1) = 0\), so the solutions are \(x = -5\) and \(x = 1\). Therefore, the x-intercepts are at \(x = -5\) and \(x = 1\).

Step 5: Find the y-intercept

The y-intercept of a function is the point where \(x = 0\). Substituting \(x = 0\) into the function gives \(f(0) = (0)^2 + 4(0) - 5 = -5\). Thus, the y-intercept is \((0, -5)\).

Step 6: Summarize graphing information

The function \(f(x) = x^2 + 4x - 5\) has the vertex \((-2, -9)\) and it opens upward. The x-intercepts are \((-5, 0)\) and \((1, 0)\), and the y-intercept is \((0, -5)\). This information, along with the vertex and intercepts, helps in graphing the parabola.

Key Concepts

Vertex of Parabola

In a quadratic function, the vertex is a crucial point as it indicates the peak or the lowest dip of the graph. For the quadratic equation \( f(x) = ax^2 + bx + c \), the vertex can be found using the vertex formula: \( h = -\frac{b}{2a} \) and \( k = f(h) \).
For the problem \( f(x) = x^2 + 4x - 5 \), we begin by identifying the coefficients: \( a = 1 \), \( b = 4 \), and \( c = -5 \). Plug these into the formula to find \( h \):

\( h = -\frac{4}{2 \times 1} = -2 \)

Next, substitute \( h = -2 \) back into the function to compute \( k \):

\( k = (-2)^2 + 4(-2) - 5 = -9 \)

Thus, the vertex of this parabola is at \( (-2, -9) \). This point informs us about the extremum of the parabola's graph.

Graphing Quadratic Equations

Graphing a quadratic function involves plotting points and understanding its shape. The basic form of a quadratic equation is \( f(x) = ax^2 + bx + c \), and its graph is a parabola. Here, we are working with a standard parabola equation.
To effectively graph the function:

• Plot the vertex \((-2, -9)\) as it indicates the graph's turning point.


• Plot the intercepts that we have calculated, which provide points where the graph crosses the axes.


• Sketch the parabola by considering its direction of opening.

Use these key points as a framework to draw the parabola, making sure the shape is symmetric around the vertex. This symmetry means that if you fold the graph along the line through the vertex, the two halves match.

Parabola Intercepts

Intercepts are essential to understanding the graph's intersections with the axes.
**X-intercepts:** These points occur where the graph crosses the x-axis, meaning \( f(x) = 0 \). Solve this equation:\[ x^2 + 4x - 5 = 0 \] Factoring gives us:

• \((x + 5)(x - 1) = 0\)


• Thus, \(x = -5\) and \(x = 1\)

These solutions provide the x-intercepts \((-5, 0)\) and \((1, 0)\).
**Y-intercept:** This is where \( x = 0 \) in the equation. Substitute it to find:
\[ f(0) = 0^2 + 4(0) - 5 = -5 \] So, the y-intercept is at \((0, -5)\). Each intercept serves as a guidepost to shape the final graph.

Direction of Parabola Opening

The direction in which a parabola opens is directly influenced by the coefficient \( a \) in the quadratic equation \( ax^2 + bx + c \).

• If \( a > 0 \), the parabola opens upward.


• If \( a < 0 \), it opens downward.

In our exercise with \( f(x) = x^2 + 4x - 5 \), the leading coefficient \( a = 1 \) is positive.
This means the parabola opens upwards, forming a 'U' shape. Understanding the direction of the opening is vital as it indicates whether the vertex is a minimum or maximum point. Since \( a > 0 \), the vertex at \( (-2, -9) \) is a minimum.