Question

The vertex of this parabola is at (2,-1). When the y value is 0, the value is 5.

Answer 1

Answer:

Option D.

Step-by-step explanation:

The vertex form of a parabola along y-axis is

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

where, (h,k) is vertex and, a is constant and it is equal to coefficient of the squared term in the parabola's equation.

The vertex of the parabola is (2,-1). So, h=2 and k=-1.

$y=a(x-2)^2-1$

The graph passes through (5,0). So,

$0=a(5-2)^2-1$

$1=9a$

$\dfrac{1}{9}=a$

It means coefficient of the squared term is 1/9, which is not the option. So, parabola must be along the x-axis.

The vertex form of a parabola along x-axis is

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

where, (h,k) is vertex and, a is constant and it is equal to coefficient of the squared term in the parabola's equation.

The vertex of the parabola is (2,-1). So, h=2 and k=-1.

$x=a(y+1)^2+2$

The graph passes through (5,0). So,

$5=a(0+1)^2+2$

$5-2=a$

$3=a$

It means coefficient of the squared term is 3.

Therefore, the correct option is D.