Problem 46 - Algebra 1

Question

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

Step-by-Step Solution

Answer

The graph of the function \(y=4x^{2}\) is a parabola that opens upward. The vertex is located at the origin, which is also the y-intercept and the x-intercept.

Step 1: Identify the Vertex

For any quadratic function in the standard form \(y=a(x-h)^{2}+k\), the vertex is at point (h, k). In this case, \(y=4x^{2}\) is already in standard form where a is 4, h is 0 and k is 0. Therefore, the vertex is at (0, 0)

Step 2: Identify the y-intercept

For the function \(y=4x^{2}\), to find the y-intercept, we set \(x=0\), then we get \(y=4*0^{2}\) which equals 0. So the y-intercept is also at (0, 0).

Step 3: Identify the x-intercepts

To find x-intercepts, set \(y=0\), then we solve the equation 0=4x^2.Weight the equation by first dividing both sides by 4 and then taking the square root of both sides will lead to x for which y equals zero. Here \(x=0\) is the x-intercept.

Step 4: Sketch the graph

Now, we know the vertex is at (0, 0), the y-intercept is also at (0, 0) and there is one x-intercept at (0, 0). Include all these points onto the graph, then sketch a parabola that opens upward since value of a is positive. Make sure the curve is smooth and symmetrical.

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function that is defined by an equation of the form



\[ y = ax^2 + bx + c \]

where

\( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). It's graphed as a curve called a parabola. These functions are characterized by the highest exponent of \( x \) being 2. The coefficient \( a \) determines the direction of the parabola, opening upwards if \( a > 0 \) and downwards if \( a < 0 \).

If we look at our exercise with the equation

\[ y = 4x^2 \]

we can see that \( b = 0 \) and \( c = 0 \), making it a simplified form of the quadratic equation. Here, \( a = 4 \), therefore, we can expect our parabola to open upwards.

Graphing Parabolas

Graphing parabolas involves plotting the curve described by a quadratic function. To begin graphing, we find key features such as the vertex, the axis of symmetry, and the intercepts.



In our example, \( y = 4x^2 \) represents a parabola with a vertex at the origin (0,0). This also means our parabola is symmetrical along the y-axis. To sketch, we start at the vertex, plot additional points on one side of the axis of symmetry, and then reflect those points across the axis to get the other side of the parabola. Remember, the more points we plot, the more accurate our graph will be.

Vertex of a Parabola

The vertex of a parabola is the highest or lowest point on the graph, depending on whether the parabola opens up or down. It's also the point where the parabola changes direction.

For a standard quadratic function, \( y = a(x - h)^2 + k \), the vertex is at \( (h, k) \). If \( a > 0 \), the vertex is a minimum point; if \( a < 0 \), it's a maximum point.

In the given problem where we have \( y = 4x^2 \), there is no \( h \) or \( k \) because the function is equivalent to \( y = 4(x - 0)^2 + 0 \), placing the vertex at the origin, \( (0, 0) \), and in this case, it is a minimum point as the \( a \) value (4) is positive.

Y-intercept

The y-intercept of a graph is the point where the graph crosses the y-axis. To find this, we set \( x = 0 \) in the quadratic equation.

For \( y = 4x^2 \), plugging in \( x = 0 \) yields \( y = 4(0)^2 = 0 \). So, the y-intercept is also at (0, 0), the same as the vertex for this particular function. This means that our graph will touch the y-axis at the origin, where it will also turn.

X-intercepts

X-intercepts are points where the graph crosses the x-axis, and to find them, we solve for \( x \) when \( y = 0 \).

In our equation \( y = 4x^2 \), setting \( y = 0 \) gives us \( 0 = 4x^2 \). Solving for \( x \), we divide both sides by 4 and get \( x^2 = 0 \). Taking the square root on both sides gives us \( x = 0 \), indicating that our function has a single x-intercept at the origin, (0,0). This point is also known as a 'double root' because the parabola just touches the x-axis at this point and does not cross it.