Question

\(y=-\frac{1}{4}(x+2)^2+1\)

Short answer

The vertex of the parabola is at (-2,1), it opens downwards and the y-intercept is 0.

Step-by-step solution

Identifying the Vertex

Looking at the given equation \(y=-\frac{1}{4}(x+2)^2+1\), we can deduce that the vertex of the parabola is at (-2, 1). This is because the vertex form of the parabolic equation is \(y=a(x-h)^2 + k\), where (h, k) is the vertex point. By comparing the given equation with the standard form, we can find h=-2 and k=1.

Determining the direction of the parabola

The sign of the coefficient 'a' in the standard form of the parabola equation determines whether the parabola opens up or down. If 'a' is positive, the parabola opens upwards and if 'a' is negative, it opens downwards. From our equation, 'a' is \(-\frac{1}{4}\) which is a negative number. So, the parabola opens downwards.

Finding the y-intercept

The y-intercept of the equation can be found by setting x to 0. So, replacing x with 0 in our equation gives \(y=-\frac{1}{4}(0+2)^2+1\), which simplifies to y=0. So 0 is the y-intercept.

Key concepts

Vertex Form

The vertex form of a quadratic equation is a special way of expressing the equation of a parabola. It allows us to quickly identify the vertex of the parabola, making it a handy tool for graph analysis. The vertex form is given by the equation:

• \(y = a(x - h)^2 + k\)

In this equation, \((h, k)\) represents the vertex of the parabola. The vertex is a significant point as it indicates the highest or lowest point of the parabola, depending on its direction of opening. In the equation \(y=-\frac{1}{4}(x+2)^2+1\), we see that \(h = -2\) and \(k = 1\). Hence, the vertex of this parabola is located at \((-2, 1)\). Understanding the vertex form not only helps in identifying the vertex but also simplifies the process of graphing the parabola.

Parabola Direction

Determining the direction in which a parabola opens is crucial for visualizing its graph. The direction is defined by the coefficient 'a' in the quadratic equation. Specifically:

• If \(a > 0\), the parabola opens upwards, creating a U-shape.
• If \(a < 0\), the parabola opens downwards, forming an inverted U-shape.

For the equation \(y=-\frac{1}{4}(x+2)^2+1\), the coefficient 'a' is \(-\frac{1}{4}\), which is negative. Therefore, the parabola opens downwards. This implies that as the x-values move away from the vertex, the y-values decrease. Recognizing the sign of 'a' helps predict the overall shape and behavior of the graph.

Y-Intercept

The y-intercept is where the graph of the equation crosses the y-axis. At this point, the value of \(x\) is zero. This makes finding the y-intercept straightforward as it can be calculated by substituting \(x = 0\) into the equation.

• For \(y=-\frac{1}{4}(x+2)^2+1\), substitute \(x = 0\):
• \(y = -\frac{1}{4}(0+2)^2+1 = -\frac{1}{4}(4)+1 = -1+1 = 0\)

Thus, the y-intercept is at \(y = 0\). This point is another essential feature of graphing because it indicates where the parabola will intersect the vertical axis. By understanding how to find and interpret the y-intercept, students can enhance their ability to sketch the graph of a quadratic equation accurately.