Question

A student is given a quadratic equation in vertex form. He determines the vertex and axis of symmetry, and then sketches the parabola so that it is symmetrical about the axis of symmetry. Is this sufficient? A: The vertex and axis of symmetry are not sufficient. without the x-intercept it is impossible to sketch the graph of a parabola accuratelyB: the vertex and axis of symmetry are sufficient. as long as the parabola's vertex is located only parabola can be graphed.C: the vertex and axis of symmetry are sufficient as long as the parabola is drawn symmetrically about its axis of symmetry only one parabola can possibly be a graphed.D: the vertex and axis of symmetry are not sufficient. Without the intercept and additional sample points falling on the parabola there are infinitely parabolas that could be drawn.

Answer 1

D

When you have the vexter, the axis of symmetry is automatially known.

But, then you only have 1 point.

A parabola, can open upwards from here, downwards, skinny, obese etc.

So, there are infinitely many parabolas that can go through that vertex.

Hence,

The vertex and axis of symmtery are not sufficient. With the intercepts (x and y) and some additional sample points falling on the parabola, there can be infinite parabolas going through the vertex.

Correct answer is D