##### multi choice question

 Mathematics Tutor: None Selected Time limit: 1 Day

May 28th, 2015

A parabola always opens upwards or downwards based on the value of a.

If a < 0, the parabola opens upwards.

If a > 0, the parabola opens downwards.

So, the lowest or the highest point(depending on the value of a) on the parabola will be the vertex.

This means the range of the parabola will never be  $(-\infty,\infty)$.

Let us take an example

Consider $f(x)=x^2+2x+3$

Vertex = (-1, 2)

The range of f(x) is all the values of y coordinates.

So, range = [2, $\infty$)

Domain of f(x) is all the values of x coordinates.

So, domain = $(-\infty,\infty)$

The x-intercepts will be the real zeros of the function. These will be the values of x for which

f(x) = 0

For the given example, it does not have any real zeros as it does not intersect the x-axis.

May 28th, 2015

The vertex form of equation of parabola is

$y=a(x-h)^2+k$

where (h , k) is the vertex of the parabola.

In the example, vertex = (-1, 2)

So, h = -1 , k = 2

The range is dependent on the  value of k and not on h.

So, option D)  is wrong.

May 28th, 2015

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May 28th, 2015
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