Quick Check Solutions Graphing Quadratics Target A
1) For \(j(x)\):
Since the \(a\) value is negative \(2\), the graph would open down and be vertically stretched. Since we are adding an h value of \(4\), and subtracting a \(k\) value of \(5\), the graph would be translated to the left \(4\) units and down \(5\) units. When sketching the graph, you want to make your parabola upside down, and more narrow than the graph of the reference function. Also, you would plot the vertex at \((-4, -5)\) instead of at the origin. The sketch of the graph would look like this:
1) For \(j(x)\):
Since the \(a\) value is negative \(2\), the graph would open down and be vertically stretched. Since we are adding an h value of \(4\), and subtracting a \(k\) value of \(5\), the graph would be translated to the left \(4\) units and down \(5\) units. When sketching the graph, you want to make your parabola upside down, and more narrow than the graph of the reference function. Also, you would plot the vertex at \((-4, -5)\) instead of at the origin. The sketch of the graph would look like this:
2) For \(k(x)\):
Since the \(a\) value is positive \(\Large\frac{1}{2}\), the graph would open up and be vertically compressed. Since we are subtracting an \(h\) value of \(3\), the graph would translate \(3\) units to the right. Since we are not adding or subtracting a \(k\) value, the graph would not translate up or down. When sketching the graph, you would draw your parabola facing upwards and slightly wider than the graph of the reference function. The vertex of the graph would be at \((3, 0)\) since you are translating horizontally to the right but not vertically. The sketch of the graph would look like this:
Since the \(a\) value is positive \(\Large\frac{1}{2}\), the graph would open up and be vertically compressed. Since we are subtracting an \(h\) value of \(3\), the graph would translate \(3\) units to the right. Since we are not adding or subtracting a \(k\) value, the graph would not translate up or down. When sketching the graph, you would draw your parabola facing upwards and slightly wider than the graph of the reference function. The vertex of the graph would be at \((3, 0)\) since you are translating horizontally to the right but not vertically. The sketch of the graph would look like this: