Problem 43 - Beginning and Intermediate Algebra

Question

Graph each function using the vertex formula. Include the intercepts. \(y=x^{2}+2 x-3\)

Step-by-Step Solution

Answer

The vertex of the given quadratic function is (-1, -4), the y-intercept is (0, -3), and the x-intercepts are (1, 0) and (-3, 0). Plot these points and draw a parabola that passes through them, opening upwards to obtain the graph of the function \(y = x^2 + 2x - 3\).

Step 1: Identify the coefficients a, b, and c

The given function is in the form of \( y = ax^2 + bx + c \). We can identify the coefficients \(a\), \(b\), and \(c\) as follows: \( a = 1, b = 2, c = -3 \).

Step 2: Find the vertex using the vertex formula

The vertex formula is \( x_v = \frac{-b}{2a} \). - Find the x-coordinate of the vertex: \( x_v = \frac{-2}{2(1)} = -1 \). - Substitute the x-coordinate into the equation to find the y-coordinate of the vertex: \( y_v = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4 \). - The vertex is (-1, -4).

Step 3: Find the y-intercept

To find the y-intercept, set \(x = 0\) and solve for \(y\). \( y = (0)^2 + 2(0) - 3 = - 3 \). The y-intercept is (0, -3).

Step 4: Find the x-intercepts

To find the x-intercepts, set \(y = 0\) and solve for \(x\). \( 0 = x^2 + 2x - 3 \). To solve this quadratic equation, we can apply the quadratic formula, which is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). - Calculate the discriminant: \( \Delta = b^2 - 4ac = 2^2 - 4(1)(-3) = 4 + 12 = 16 \). - Substitute the values of \(a\), \(b\), and \(\Delta\) into the quadratic formula: \( x = \frac{-2 \pm \sqrt{16}}{2} \). - Solve for x: \( x_1 = \frac{-2 + 4}{2} = 1 \) and \( x_2 = \frac{-2 - 4}{2} = -3 \). The x-intercepts are (1, 0) and (-3, 0).

Step 6: Sketch the graph

Now that we have the vertex, x-intercepts, and y-intercept, we can sketch the graph of the given quadratic function: 1. Plot the vertex (-1, -4). 2. Plot the x-intercepts (1, 0) and (-3, 0). 3. Plot the y-intercept (0, -3). 4. Draw a parabola that passes through the plotted points, opening upwards (since the coefficient of the x^2 term is positive). After plotting these points and drawing the parabola, you will have the graph of the given quadratic function, \( y = x^2 + 2x - 3 \).

Key Concepts

Vertex Formula

The vertex of a parabola is a crucial point since it represents either the highest or lowest point of the graph, depending on the direction the parabola opens. To find the vertex, we use the vertex formula, which provides the x-coordinate of the vertex.

• The formula is: \( x_v = \frac{-b}{2a} \), where \( b \) and \( a \) are coefficients from the quadratic equation \( ax^2 + bx + c \).
• After finding \( x_v \), plug it back into the quadratic equation to find the y-coordinate \( y_v \).
• In our example \( y = x^2 + 2x - 3 \), the coefficients are \( a = 1 \) and \( b = 2 \).
• The x-coordinate of the vertex is: \( x_v = \frac{-2}{2 \, \times \, 1} = -1 \).
• The y-coordinate, by substituting \( x_v = -1 \) back into the equation: \( y_v = (-1)^2 + 2(-1) - 3 = -4 \).

Thus, the vertex of the parabola is \((-1, -4)\). This calculation confirms the turning point of the quadratic function.

X-intercepts

X-intercepts are the points where the graph intersects the x-axis. At these points, the value of \( y \) is zero. Finding the x-intercepts requires solving the quadratic equation set to zero.

• The general formula for finding x-intercepts is setting \( y = 0 \), thus \( ax^2 + bx + c = 0 \).
• We use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( \pm \) indicates two solutions.
• Calculate the discriminant \( \Delta = b^2 - 4ac \). If \( \Delta \) is positive, there are two real x-intercepts.
• For our equation \( x^2 + 2x - 3 \): \( \Delta = 2^2 - 4 \cdot 1 \cdot (-3) = 16 \), positive, indicating two intercepts.
• Substituting into the quadratic formula: \( x_1 = \frac{-2 + 4}{2} = 1 \) and \( x_2 = \frac{-2 - 4}{2} = -3 \).

Therefore, the x-intercepts are \((1, 0)\) and \((-3, 0)\). These are the points where the parabola crosses the x-axis.

Y-intercept

The y-intercept is the point where the graph crosses the y-axis. At this point, the value of \( x \) is zero. This is typically the easiest intercept to find.

• Simply substitute \( x = 0 \) in the quadratic equation \( y = ax^2 + bx + c \).
• For the function \( y = x^2 + 2x - 3 \), plug \( x = 0 \):
• \( y = (0)^2 + 2(0) - 3 = -3 \).
• Hence, the y-intercept is \((0, -3)\).

This intercept gives a crucial point for graphing the function, representing where it meets the y-axis.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It is particularly useful when factorizing is not straightforward.

• This formula is: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula gives two solutions corresponding to the x-intercepts of a parabola.
• The \( \pm \) part indicates two possible values for \( x \), leading to two intercepts.
• The term \( b^2 - 4ac \), known as the discriminant, determines the number and type of solutions.
• If the discriminant is positive, there are two distinct real solutions (x-intercepts).
• In our example, \( \Delta = 16 \), hence the two solutions are x-intercepts: \((1, 0)\) and \((-3, 0)\).

Remember, the quadratic formula is essential for solving quadratics and understanding the nature of their graphs.