Problem 54 - Algebra 1

Question

SKETCHING GRAPHS Sketch the graph of the function. Label the vertex. $$ y=x^{2}+x+\frac{1}{4} $$

Step-by-Step Solution

Answer

The vertex of the function \(y=x^{2}+x+\frac{1}{4}\) is \((-\frac{1}{2}, -\frac{1}{16})\). It is an upward-opening parabola.

Step 1: Identify the coefficients

From the given function \(y=x^{2}+x+\frac{1}{4}\), the coefficients \(a\), \(b\), and \(c\) are respectively 1, 1, and \(\frac{1}{4}\).

Step 2: Calculate the vertex

Using the formula \((-\frac{b}{2a}, f(-\frac{b}{2a})\), calculate the x-coordinate of the vertex. Substitute \(a=1\), \(b=1\) into \(-\frac{b}{2a}\) to get \(-\frac{1}{2}\). Then find the y-coordinate by substituting \(x=-\frac{1}{2}\) into the function \(y=x^{2}+x+\frac{1}{4}\).

Step 3: Sketch the graph

This is an upward-opening parabola because \(a=1\) is positive. Plot the vertex point calculated in Step 2. Draw the axis of symmetry, the vertical line \(x=-\frac{1}{2}\). Since the parabola is symmetric about this line, plot other points to the right of the vertex and reflect them in the axis of symmetry to find points on the left of the vertex. Once there are enough points, sketch the curves through those points to get the graph of the function.

Key Concepts

Sketching Parabolas

Learning to sketch parabolas is an essential skill in understanding quadratic functions. The shape of a parabola is like a U (if it opens upwards) or an inverted U (if it opens downwards), depending on the sign of the leading coefficient. For the function
\( y = x^2 + x + \frac{1}{4} \),
the leading coefficient is positive, indicating it will open upwards.

To start sketching, first plot the vertex, which is the highest or lowest point on the graph depending on the parabola's direction. Here, it's at \( (-\frac{b}{2a}, f(-\frac{b}{2a})) \). You can then identify a few points around the vertex and plot them. Remember, the parabola is symmetrical about its axis, so points equidistant from the axis on either side will have the same y-value. Once enough points are plotted, draw a smooth curve through them to complete your sketch. It might help to use graph paper and a ruler, ensuring the parabola's branches are symmetrical and smooth.

Vertex of a Parabola

The vertex of a parabola is quite literally its 'turning point'. For the function in our exercise, the vertex is found using the formula \( (-\frac{b}{2a}, f(-\frac{b}{2a})) \). Here, we substitute \( a=1 \) and \( b=1 \) into the formula to determine the x-coordinate of the vertex is \( -\frac{1}{2} \).

To find the y-coordinate, we plug \( x=-\frac{1}{2} \) back into the quadratic function, resulting in the y-coordinate. So, the vertex for our graph is \( (-\frac{1}{2}, y-value) \). This point serves as a crucial reference for drawing the parabola since it helps you understand the shape and position of the graph on the coordinate plane.

Axis of Symmetry

The axis of symmetry in a parabola is a vertical line that perfectly divides the parabola into two mirror-image halves. It’s always a straight line that passes through the vertex. The general formula for the axis of symmetry of a quadratic function \( y = ax^2 + bx + c \) is \( x = -\frac{b}{2a} \).

For our function, substituting \( a \) and \( b \) into this formula gives us an axis of symmetry at \( x = -\frac{1}{2} \). This axis is not only central to sketching a balanced parabola but also to analyzing the parabola's properties—like determining if points on one side will have corresponding points with equal y-values on the other side. When graphing, draw the axis of symmetry first to guide you in finding and plotting points that will accurately shape your parabola.