Problem 56 - Algebra 1

Question

SKETCHING GRAPHS Sketch the graph of the function. Label the vertex. $$ y=2 x^{2}+6 x-5 $$

Step-by-Step Solution

Answer

The vertex of the function \(y=2 x^{2}+6 x-5\) is at \((-1.5, -5.5)\). Sketching this on a graph results in a parabola with the vertex at \((-1.5,-5.5)\), opening upwards and crossing the y-axis at \(y = -5\).

Step 1: Determine the vertex

First, write down the quadratic function. Here: \( y = 2x^2 + 6x - 5 \). For a function of the form \( y = ax^2 + bx + c \), the vertex can be found using the formula \(h = -b/2a\). In this case, \(a = 2, b = 6\). Thus, the x-coordinate of the vertex, \(h\), is \(-6/(2*2) = -1.5\). To find the y-coordinate, \(k\), substitute \(h\) into the equation: \( y = 2(-1.5)^2 + 6(-1.5) - 5 = -5.5 \). Therefore, the vertex is \((-1.5, -5.5)\).

Step 2: Plot the vertex and the parabola

Now that the vertex has been found, you can plot this point on a coordinate plane. Given this is a parabola, it’s known that it will be symmetric about the line \(x = h\), and it will open upwards because \(a > 0\). That means the parabola will be lowest at the vertex. The y-intercept can also be found by setting \(x = 0\) in the original equation, which results in \(y = -5\). You have two points now: the vertex and the y-intercept. Plot these points, sketch the axis of symmetry at \(x = h\), and complete the graph using the symmetry of the parabola.

Key Concepts

Graphing Parabolas

When graphing a quadratic function like the one given by \(y = 2x^2 + 6x - 5\), it is essential to understand its shape. A quadratic function graphs into a parabola, which looks like a U-shape. The direction in which the parabola opens depends on the coefficient \(a\) of \(x^2\).

- If \(a > 0\), the parabola opens upwards.- If \(a < 0\), the parabola opens downwards.
For our function, \(a = 2\), meaning it opens upwards. To sketch the parabola efficiently, identify key points such as the vertex and intercepts, and understand the symmetry to ensure the parabola is accurate. Begin by plotting these key points, then draw a symmetric curve through them.

Vertex Form

The vertex form of a quadratic function is a handy form, as it directly shows the vertex of the parabola. The standard vertex form is expressed as \(y = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola.

With the given function \(y = 2x^2 + 6x - 5\), we don't initially have it in vertex form. However, you can convert it using the formula for finding the vertex \(h = -b/(2a)\). After calculating \(h = -1.5\), substitute back into the function to find \(k\), which yields \(k = -5.5\). Thus, the vertex is \((-1.5, -5.5)\).

Knowing the vertex helps immensely in graphing as it represents the highest or lowest point of the parabola, depending on its direction.

Symmetry of Parabolas

Parabolas have a line of symmetry that passes through their vertex. This line is essential because it means one side of the parabola is a mirror image of the other. For any quadratic function of the form \(y = ax^2 + bx + c\), the axis of symmetry can be expressed as \(x = h\), where \(h\) is the x-coordinate of the vertex.

For our example, since \(h = -1.5\), the axis of symmetry is \(x = -1.5\). Drawing this line helps in accurately sketching the parabola by ensuring the curve is equidistant from both sides. The symmetry aids in predicting other points and completing the graph effectively.

Finding Intercepts

Intercepts are crucial for sketching a graph. They indicate where the graph crosses the axes. For a quadratic function:

• The y-intercept is found by setting \(x = 0\) in the equation and solving for \(y\). In our case, substituting \(x = 0\), you get \(y = -5\). So, the y-intercept is the point \((0, -5)\).
• The x-intercepts (if they exist) are found by setting \(y = 0\) and solving the quadratic equation for \(x\). This involves using the quadratic formula or factoring the equation, whichever is applicable.

In the equation \(y = 2x^2 + 6x - 5\), solving for x-intercepts means setting \(2x^2 + 6x - 5 = 0\). Using the quadratic formula gives potential x-intercept points, helping complete the graph.

These intercepts provide plotting guidance, marking where the parabola crosses the axes, and complement insights gathered from the vertex and symmetry.