Problem 42 - Algebra 1: Concepts and Skills

Question

Sketch the graph of the function. Label the coordinates of the vertex. $$ y=2 x^{2}+5 x-3 $$

Step-by-Step Solution

Answer

The vertex of the function \(y=2x^{2}+5x-3\) is at \((-5/4, -7)\). The graph is an upward-opening parabola that passes through this vertex.

Step 1: Identify the Form of the Function

The given function \(y=2x^{2}+5x-3\) is in the general form of a quadratic function \(y=ax^{2}+bx+c\). The coefficients are \(a=2\), \(b=5\), and \(c=-3\). Since \(a>0\), the graph of the function is a parabola that opens upwards.

Step 2: Find the Vertex

The vertex of a parabola \(y=ax^{2}+bx+c\) is at \(h=-\frac{b}{2a}\). Substituting \(b=5\) and \(a=2\) we find \(h=-\frac{5}{4}\). Substituting \(h=-\frac{5}{4}\) into the equation, we find \(k=2(-\frac{5}{4})^{2}+5(-\frac{5}{4})-3=-7\). Therefore, the vertex is \((-5/4, -7)\).

Step 3: Sketch the Graph

Draw the x and y axes. Plot the vertex point \((-5/4, -7)\). Sketch an upward-opening parabola through the plotted vertex.

Key Concepts

Vertex of a Parabola

When graphing a quadratic function, a key feature to locate is the vertex of the parabola. The vertex is the highest or lowest point on the graph, known as the maximum or minimum respectively. This point is crucial because it represents the turning point of the parabola where the function changes direction. For a function in the form of
\(y = ax^{2} + bx + c\),
you can find the vertex with the formula
\(h = -\frac{b}{2a}\). Once you compute the value of \(h\), you can plug it back into the function to find the \(k\) value, giving you the vertex coordinates \((h, k)\).
In the provided exercise, the vertex of the parabola \(y = 2x^{2} + 5x - 3\),
was found by substituting \(a = 2\) and \(b = 5\) into the vertex formula, yielding \(h = -\frac{5}{4}\). Following this, \(k\) was computed as \(-7\), thus giving the vertex coordinates of \((-\frac{5}{4}, -7)\). Identifying the vertex helps you understand the parabola's direction and provides a starting point for graphing.

Quadratic Function

A quadratic function is a second-degree polynomial expressed in the form
\(y = ax^{2} + bx + c\),
where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The graph of a quadratic function is a parabola. If \(a > 0\), the parabola opens upwards --- as in the example problem --- and if \(a < 0\), it opens downwards. The value of \(a\) affects the width of the parabola; larger values make it narrower while smaller values make it wider. The coefficient \(b\) affects the position of the vertex along the horizontal axis, and \(c\) represents the y-intercept, or the point where the graph intersects the y-axis. Understanding these components is valuable when sketching the graph as it outlines the shape and position of the parabola in the coordinate plane.

Parabola Sketching

Sketching a parabola requires understanding its key features, including the vertex, axis of symmetry, intercepts, and the direction it opens. After finding the vertex, as we have in the exercise, the next step is to draw the axis of symmetry, a vertical line that passes through the vertex. This axis divides the parabola into two mirror images. Subsequently, you can find the y-intercept, which is simply the value of \(c\) in the quadratic function, and plot that on the graph.
To have more points for accuracy, you can also calculate the x-intercepts, if they exist, by setting \(y = 0\) and solving for \(x\). Finally, consider the value of \(a\) for the direction and width: an upward-opening parabola for \(a > 0\) and wider for smaller values of \(a\). With these points and the knowledge of the parabola's direction, draw a smooth curve through the vertex and intercepts to complete the sketch.
Remembering these steps simplifies the process of parabola sketching, turning it into an achievable task following a methodical approach.