Question

Which quadratic function opens upwards and has a vertex at ( 0 , 3 )?

A.) y = -(x - 3)^2

B.) y = (x - 3)^2

C.) y = -x^2 + 3

D.) y = x^2 + 3

Answer 1

Answer:

the vertex is at (0,3). Notice that in this problem the vertex and the y-intercept are the same point. Since a 0 the parabola opens up (is U shaped).

Step-by-step explanation:

To find the x-intercepts we plug in 0 for y:

0 = -3x2 + 3 (this equation factors)

0 = -3(x2 - 1)

0 = -3(x - 1)(x + 1) and since -3 can not equal zero:

x = 1 or x = -1

The x-intercepts are: (1,0) and (-1,0)

The y-intercept is found by plugging 0 for x:

y = -3(0)2 + 3 = 3

So, the y-intercept is at (0,3).

And to find the vertex:

k = -3(0)2 + 3 = 3

So the vertex is at (0,3).

Notice that in this problem the vertex and the y-intercept are the same point.