Problem 16 - Algebra 1

Question

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

Step-by-Step Solution

Answer

The graph of function \(y=5x^2+5x-2\) is a parabola with vertex at (-0.5, -0.5) which opens upwards.

Step 1: Find the vertex

The vertex is found using the formula \(h=-\frac{b}{2a}\) and \(k=f(h)\). Here, a=5, b=5. Substituting these values gives \(h=-\frac{5}{2*5}=-\frac{1}{2}\). Obtain k by substituting h into the equation where we get \(k=5(-\frac{1}{2})^{2}+5(-\frac{1}{2})-2= -\frac{1}{2}\). Therefore, the vertex is (-0.5, -0.5).

Step 2: Determine the direction of the graph

The parabola will open upwards if 'a' is positive and it will open downwards if 'a' is negative. In our case, a is 5 which is positive so the parabola opens upwards.

Step 3: Sketch the graph

Plot the vertex that was found in step 1 on the graph. Next, because the parabola opens upwards as determined in step 2, sketch the graph with the vertex at (-0.5, -0.5) and the graph opening upwards.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial feature in the graph of a quadratic function. It represents the highest or lowest point on the graph, also known as either a maximum or a minimum, depending on the direction in which the parabola opens. To find the vertex, we use the vertex formula where \( h = -\frac{b}{2a} \). This formula helps locate the x-coordinate of the vertex when a quadratic function is given in the standard form \( y = ax^2 + bx + c \). In our function, \( y = 5x^2 + 5x - 2 \), we plug in the values of \( a = 5 \) and \( b = 5 \) into the formula, resulting in \( h = -\frac{1}{2} \).
Once the x-coordinate is determined, find the y-coordinate \( k = f(h) \) by substituting \( h \) back into the original quadratic equation. For our function, substituting \( h = -\frac{1}{2} \) back, we find \( k = 5(-\frac{1}{2})^2 + 5(-\frac{1}{2}) - 2 = -\frac{1}{2} \). Thus, the vertex of this quadratic function is \((-0.5, -0.5)\), serving as a keystone for graphing the entire parabola.

Graphing Quadratic Equations

A quadratic equation can be graphed as a smooth U-shaped curve known as a parabola. Graphing these equations involves certain steps which help accurately render the curve. The vertex, which is the pivotal point, needs to be plotted first since it determines the center of the parabola.
After finding and sketching the vertex, as we did for \( y = 5x^2 + 5x - 2 \) with the vertex \((-0.5, -0.5)\), the next step is to determine the direction in which the parabola opens. Once this is settled, the parabola can be sketched symmetrically about the vertical line that passes through the vertex. In some cases, calculating additional points by substituting other x-values into the quadratic function can help in creating a more accurate sketch. This technique ensures the curve is smooth and correctly shaped. Quadratic functions always yield parabolas, making the process of graphing these equations consistently recognizable.

Parabola Opening Direction

The direction in which a parabola opens is dictated by the coefficient \( a \) in the quadratic equation of the form \( y = ax^2 + bx + c \).
Here’s how it works:

• If \( a > 0 \), the parabola opens upwards like a smile.
• If \( a < 0 \), the parabola opens downwards like a frown.

In our specific function \( y = 5x^2 + 5x - 2 \), because \( a = 5 \) is positive, the parabola opens upwards. This direction affects not only the graphical appearance of the function but also whether the vertex represents a minimum or maximum point. When the parabola opens upward, as in this case, the vertex \((-0.5, -0.5)\) is at the lowest point of the curve, representing a minimum. Understanding this concept is fundamental for correctly graphing and interpreting quadratic functions.