Problem 52 - Beginning and Intermediate Algebra

Question

Graph each function using the vertex formula. Include the intercepts. \(y=\frac{1}{2} x^{2}+2 x-3\)

Step-by-Step Solution

Answer

The vertex of the function \(y=\frac{1}{2}x^2 + 2x - 3\) is (-1, -4), and the axis of symmetry is \(x=-1\). The y-intercept is (0, -3), and the x-intercepts are (\(-1 - \sqrt{7}\), 0) and (\(-1 + \sqrt{7}\), 0). Plot these points and the axis of symmetry, and sketch the parabola opening upwards.

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

From the given quadratic function, the coefficients are: \(a=\frac{1}{2}\), \(b=2\), and \(c=-3\)

Step 2: Find the vertex using the formula

The vertex formula is given by: \[h = \frac{-b}{2a}\] Plugging in the values of a and b: \[h = \frac{-2}{2 \times \frac{1}{2}}\] \[h = -1\] Now, to find the y-coordinate (k) of the vertex, we can plug h back into the original equation: \[k = \frac{1}{2}(-1)^2 + 2(-1) - 3\] \[k = -4\] So, the vertex is (-1, -4).

Step 3: Determine the axis of symmetry

As the axis of symmetry is the x-coordinate of the vertex, we have: Axis of symmetry: \(x=-1\)

Step 4: Find the y-intercept

To find the y-intercept, set x to 0 in the function: \[y = \frac{1}{2}(0)^2 + 2(0) - 3\] \[y = -3\] So, the y-intercept is (0, -3).

Step 5: Find the x-intercepts

To find the x-intercepts, set y to 0 in the function: \[0 = \frac{1}{2}x^2 + 2x - 3\] We can use the quadratic formula to solve for x: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Plugging in the values of a, b, and c: \[x = \frac{-2 \pm \sqrt{2^2 - 4 \times \frac{1}{2} \times (-3)}}{2 \times \frac{1}{2}}\] \[x = -1 \pm \sqrt{7}\] So, the x-intercepts are (\(-1 - \sqrt{7}\), 0) and (\(-1 + \sqrt{7}\), 0).

Step 6: Graph the function

Plot the vertex, axis of symmetry, and the intercepts on the coordinate plane: 1. Plot the vertex (-1, -4). 2. Draw the axis of symmetry (vertical line) at x=-1. 3. Plot the y-intercept (0, -3). 4. Plot the x-intercepts (\(-1 - \sqrt{7}\), 0) and (\(-1 + \sqrt{7}\), 0). 5. Sketch the parabola opening upwards passing through these points. Now, you have drawn the graph of the given function \(y=\frac{1}{2}x^2 + 2x - 3\), including the intercepts.

Key Concepts

Vertex Formula

To graph a quadratic function like \(y = \frac{1}{2}x^2 + 2x - 3\), understanding the vertex formula is essential. The vertex formula is used to find the vertex of a parabola given by \((h, k)\). The x-coordinate \(h\) of the vertex can be determined using the formula:\[h = \frac{-b}{2a}\]where \(a\) and \(b\) are coefficients from the quadratic equation in the standard form \(ax^2 + bx + c\). By substituting the values \(a = \frac{1}{2}\) and \(b = 2\), we get:- \(h = \frac{-2}{2 \times \frac{1}{2}} = -1\)To find the y-coordinate \(k\), substitute back into the quadratic equation:- \(k = \frac{1}{2}(-1)^2 + 2(-1) - 3 = -4\)Therefore, the vertex of the parabola is \((-1, -4)\). Finding the vertex helps determine the shape and position of the parabola on a graph.

Intercepts of a Parabola

Intercepts play a crucial role in graphing and interpreting quadratic functions. They are the points where the parabola crosses the axes.

• Y-Intercept: The y-intercept is found by setting \(x = 0\) in the quadratic equation. For this function: \[ y = \frac{1}{2}(0)^2 + 2(0) - 3 = -3 \] Thus, the y-intercept is \((0, -3)\), which is the point where the graph intersects the y-axis.
• X-Intercepts: To determine the x-intercepts, set \(y = 0\) in the equation. The quadratic formula solves for \(x\): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting \(a = \frac{1}{2}\), \(b = 2\), and \(c = -3\) gives: \[ x = \frac{-2 \pm \sqrt{2^2 - 4 \times \frac{1}{2} \times (-3)}}{2 \times \frac{1}{2}} = -1 \pm \sqrt{7} \] Therefore, the x-intercepts are \((-1 - \sqrt{7}, 0)\) and \((-1 + \sqrt{7}, 0)\).

Both intercepts provide critical insight into where the function satisfies specific conditions along the axes.

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. For any quadratic equation \(y = ax^2 + bx + c\), the axis of symmetry can be easily found using the x-coordinate of the vertex \(h\), calculated by the vertex formula:\[ x = \frac{-b}{2a} \]For our specific function, we already calculated \(h = -1\). Thus, the axis of symmetry is the line:\[ x = -1 \]This line passes through the vertex \((-1, -4)\) and indicates that for every point \((x, y)\) on the graph, there is a corresponding point \((-x, y)\) equidistant on the opposite side of this line. When sketching the parabola, this line helps in drawing symmetrical curves on both sides, ensuring accuracy and balance.

Quadratic Formula

The quadratic formula is a tool used to find the roots of a quadratic equation \( ax^2 + bx + c = 0 \). These roots correspond to the x-intercepts of the parabola. The quadratic formula is written as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula calculates the solutions for \(x\) where the function equals zero (i.e., the parabola crosses the x-axis). For the equation \( y = \frac{1}{2}x^2 + 2x - 3 \), substituting the coefficients \(a = \frac{1}{2}\), \(b = 2\), and \(c = -3\) gives:\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \times \frac{1}{2} \times (-3)}}{2 \times \frac{1}{2}} \]After calculations, the solutions are: \(x = -1 \pm \sqrt{7}\). These roots or x-intercepts confirm where the graph of the quadratic function touches or crosses the x-axis. Understanding how to apply the quadratic formula is vital for solving and graphing quadratic equations.