Question

SKETCHING GRAPHS Sketch the graph of the function. Label the vertex. $$ y=3 x^{2}-2 x-1 $$

Short answer

The vertex of the function \(y=3 x^{2}-2 x-1\) is at (1/3, -4/3) and the graph opens upwards.

Step-by-step solution

Compute the vertex

To find the vertex we need to use the formula \(-b/2a , f(-b/2a)\). For this function, \(a = 3\) and \(b = -2\). Plug these values into the vertex formula to get the point which will yield: \(x = -(-2) / (2*3) = 1/3\) and \(y = 3*(1/3)^2 - 2*(1/3) - 1 = -4/3\). This means that the vertex of the function is (1/3, -4/3).

Determine the graph direction

This will tell us whether the parabola opens up or down. The sign of 'a', which is the coefficient of \(x^2\), decides the direction of the graph. If 'a' is positive, the graph opens upwards; if 'a' is negative, the graph opens downwards. Thus, because 'a' is 3 in this case (a positive number), the graph will open upwards.

Plot the graph

First plot the vertex point (1/3, -4/3). Since it opens upwards, plot additional points upwards from the vertex. You can also find points when \(x=0\) and when \(x=1\) to have more reference points and plot them too. Draw smooth curves to connect these points, ensuring it opens upwards. This will result in the desired plot of the equation \(y=3 x^{2}-2 x-1\).

Key concepts

Vertex of a Parabola

The vertex of a parabola is a critical concept when sketching quadratic graphs. It represents the highest or lowest point on the graph, depending on the parabola's orientation. To find the vertex, you can use the formula \( -\frac{b}{2a}, f(-\frac{b}{2a}) \), where \( a \) and \( b \) are coefficients from the quadratic equation in the standard form \( y = ax^2 + bx + c \).

For example, in the exercise \( y=3x^{2}-2x-1 \), we identify that \( a = 3 \) and \( b = -2 \). Applying these to the formula gives us the x-coordinate of the vertex as \( x = \frac{-(-2)}{2\times3} = \frac{1}{3} \). To find the y-coordinate, substitute \( x \) back into the original equation to get \( y = 3(\frac{1}{3})^2 - 2(\frac{1}{3}) - 1 = -\frac{4}{3} \). Consequently, the vertex of this parabola is at the point \( (\frac{1}{3}, -\frac{4}{3}) \).

Understanding the vertex is essential because it serves as the compass for the rest of the graph—highlighting symmetries and providing a starting point for plotting other points.

Parabola Graph Direction

The direction a parabola graph takes is determined by the sign of the leading coefficient—the \( a \) term in the quadratic equation \( y = ax^2 + bx + c \) . If \( a \) is positive, the parabola opens upwards, forming a 'U' shape. Conversely, if \( a \) is negative, the parabola opens downwards, resembling an upside-down 'U'.

For the exercise function \( y=3x^{2}-2x-1 \) , since \( a \) is 3, which is a positive value, we can confirm that the graph opens upwards. This direction affects the values of \( y \) as \( x \) increases or decreases: as we move away from the vertex on the x-axis, the \( y \) values increase, indicating that the arms of the parabola stretch upward. Understanding the direction is pivotal for predicting the general shape of the graph and for determining the potential range of \( y \) values for the function.

Plotting Parabolas

Plotting parabolas effectively requires both the vertex and the direction of the graph. Starting with the vertex, as identified in the previous sections, you plot this point first. For the given equation \( y=3x^{2}-2x-1 \) , that's the point \( (\frac{1}{3}, -\frac{4}{3}) \).

Once the vertex is in place, you plot additional points to form the shape of the parabola. Using the graph direction, and knowing this particular parabola opens upwards, choose x-values around the vertex, like \( x=0 \) or \( x=1 \) , and calculate the corresponding y-values by substituting them into the equation. After calculating a few points, plot them on the graph, and draw a smooth curve through the points, ensuring it opens in the correct direction.

Tips for Effective Plotting

• Use symmetry about the vertex to find pairs of points on the graph.
• Look for x-intercepts by setting \( y = 0 \) and solving for \( x \) to add more reference points.
• Ensure your curve is smooth and continuous without sharp bends.

Plotting additional points and considering the parabola's symmetry will help build an accurate and complete graph of the quadratic function.