Question

Liesa wants to graph k(x) = 2(x + 3)^2 - 5. She thinks that the vertex of the parabola should be (6,-5), because 2*3=6, and there’s a (-5) at the end. She plots that vertex, and then graphs a parent function upward from there. She ends up with the graph shown. Explain whether you agree or disagree with Liesa’s work, and what you would have done differently, if you disagree. Answer specifically about the vertex, y-intercept, and the shape and orientation of the parabola.

Please provide your analysis, and whether you agree or disagree with Liesa’s approach to graphing this function."

Answer 1

I disagree with Liesa's approach to graphing the function. The correct vertex is (-1.5, -0.5). The parabola is concave up.

Step-by-step explanation:

I disagree with Liesa's approach to graphing the function. To find the vertex of the parabola, we need to use the formula h = -b/2a, where h is the x-coordinate of the vertex. In the equation k(x) = 2(x + 3)^2 - 5, we have a = 2 and b = 6. Plugging in these values, we get h = -6/(2*2) = -1.5. So the x-coordinate of the vertex is -1.5, not 6. To find the y-coordinate of the vertex, we substitute the x-coordinate into the function: k(-1.5) = 2(-1.5 + 3)^2 - 5 = 2(1.5)^2 - 5 = 2(2.25) - 5 = 4.5 - 5 = -0.5. Therefore, the correct vertex of the parabola is (-1.5, -0.5).

The y-intercept of the function can be found by setting x = 0 and solving for y. Plugging in x = 0 into the function, we get k(0) = 2(0 + 3)^2 - 5 = 2(3)^2 - 5 = 2(9) - 5 = 18 - 5 = 13. Therefore, the y-intercept is (0, 13).

The parabola represented by the function k(x) = 2(x + 3)^2 - 5 is a U-shaped, or concave up, parabola. This means that the coefficient of x^2, which is 2, is positive. If the coefficient were negative, the parabola would be concave down. The parabola opens upwards because the y-values increase as x increases.