Question

challenge #4
plot a parabola through the points.
when you finish, press \check my work\.
(there is a coordinate grid with a vertex marked at (4, -2) (approximate), a point at (0, 6) (approximate), and a point at (6, 0) (approximate), and a text box and a \check my work\ button on the right side.)

Explanation:

Step1: Identify the vertex

The vertex of the parabola is at \((4, -2)\) (from the graph, the vertex is marked at \(x = 4\), \(y=-2\)). The standard form of a parabola is \(y=a(x - h)^2 + k\), where \((h,k)\) is the vertex. So \(h = 4\), \(k=-2\), so the equation becomes \(y=a(x - 4)^2-2\).

Step2: Find a point on the parabola

We have a point on the y - axis: when \(x = 0\), \(y = 6\) (from the graph, the point on the y - axis is \((0,6)\)). Substitute \(x = 0\), \(y = 6\) into the equation \(y=a(x - 4)^2-2\):
\[

$$\begin{align*} 6&=a(0 - 4)^2-2\\ 6&=16a-2\\ 16a&=6 + 2\\ 16a&=8\\ a&=\frac{8}{16}\\ a&=\frac{1}{2} \end{align*}$$

\]

Step3: Write the equation of the parabola

Substitute \(a=\frac{1}{2}\), \(h = 4\), \(k=-2\) into the standard form \(y=a(x - h)^2 + k\). We get \(y=\frac{1}{2}(x - 4)^2-2\). We can also expand this:
\[

$$\begin{align*} y&=\frac{1}{2}(x^{2}-8x + 16)-2\\ y&=\frac{1}{2}x^{2}-4x+8 - 2\\ y&=\frac{1}{2}x^{2}-4x + 6 \end{align*}$$

\]

To plot the parabola, we know the vertex is \((4,-2)\), the y - intercept is \((0,6)\) (from the point on the y - axis), and we can find another point. Let's use the axis of symmetry \(x = 4\). The point \((0,6)\) is 4 units left of the axis of symmetry, so the point 4 units right of the axis of symmetry (\(x=4 + 4=8\)) will have the same y - value. So when \(x = 8\), \(y = 6\) (we can check with the equation: \(y=\frac{1}{2}(8 - 4)^2-2=\frac{1}{2}(16)-2=8 - 2 = 6\)). Now we can plot the vertex \((4,-2)\), the points \((0,6)\) and \((8,6)\) and draw the parabola through these points.

Answer:

To plot the parabola, use the vertex \((4, -2)\), the y - intercept \((0, 6)\) and the point \((8, 6)\) (found using the axis of symmetry \(x = 4\)) and draw a smooth curve through these points. The equation of the parabola is \(y=\frac{1}{2}(x - 4)^2-2\) (or \(y=\frac{1}{2}x^{2}-4x + 6\)).