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viva [34]
4 years ago
9

A student is given a quadratic equation in vertex form. He determines the vertex and axis of symmetry, and then sketches the par

abola so that it is symmetrical about the axis of symmetry. Is this sufficient? Explain.
Mathematics
1 answer:
igor_vitrenko [27]4 years ago
5 0

Answer:

The information with which the parabola is constructed is insufficient

Step-by-step explanation:

The vertex form of a quadratic equation is given as follows;

f(x) = a(x - h)² + k

The steps to draw the graph of a quadratic function given in vertex form, we are required to do the following;

1) Definition of the variables, (a, h and k) of the quadratic equation (given in vertex form)

The vertex of the quadratic equation in vertex form is (h, k)

2) Given that the vertex point is given, the vertex point is then plotted on the graph

3) Draw the parabola axis of symmetry that passes through the vertex point

4) Determine the direction to which the parabola opens from the sign of the variable a as follows;

i) When a is positive, the parabola opens upwards

ii) When a is negative, the parabola opens downwards

5) Determine if the parabola has x-intercepts and plot them

A parabola has x intercepts when;

i) The parabola opens downwards, that is, the variable, a, is negative and also the vertex (h, k) is above the x-axis or k is positive

ii) The parabola opens upwards, that is, the variable, a, is positive and also the vertex (h, k) is below the x-axis or k is negative

6)  Determine y-intercepts and plot them as follows;

At the y-intercept, x = 0, therefore;

f(x) = a(x - h)² + k

f(0) = a(0 - h)² + k = a·h² + k

6) If any additional points are required, they should plotted and then the graph of the function is then drawn making sure to be symmetrical about the vertex and to pass through all the plotted points.

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