Question

Consider the quadratic function y = (x – 2)2 + 5. Which statements are true about the function and its graph? Check all that apply

The vertex of the function is (–2, 5).
There are no real roots for the function.
The graph of the function opens down.
The graph contains the point (2, 5).
The graph intersects the x-axis at one unique point.

Answer 1

The vertex of the function is (–2, 5).
There are no real roots for the function.
The graph of the function opens down.
The graph contains the point (2, 5).
The graph intersects the x-axis at one unique point.

Answer 2

There are no real roots for the function.

The graph contains the point (2, 5).

Explanation:

Let's check every option in order to conclude which of them apply:

The vertex of the function is (–2, 5).

The x-coordinate of the vertex can be found by the formula:

`(-b)/(2a)`

But, first, let's rewrite the equation in its standard form:

`y=ax^2+bx+c`

So:

`y=x^2+4x+9`

In this case:

`a=1\\b=4\\c=9`

The x-coordinate of the vertex is:

`(-b)/(2a)=(-(-4))/(2(1)) =(4)/(2) =2`

and The y-coordinate of the vertex is:

`y(2)=(2)^2-4(2)+9=4-8+9=5`

So the vertex of the function is (2, 5). Hence this statement is wrong.

There are no real roots for the function.

Use quadratic formula to find the roots:

`x=(-b\pm √(b^2-4ac) )/(2a)`

`x=(-(-4)\pm √(16-36) )/(2)\\\\x=(4\pm √(-20) )/(2)\\\\x=(4\pm √(4)*√(5)*√(-1) )/(2)`

Where:

`√(-1) =i`

`x=(4\pm 2*√(5)*i )/(2)`

So, the roots are complex and they are:

`x=2-√(5)i\\ x=2+√(5)i`

Hence this statement is true.

The graph of the function opens down.

Analyze the function:

The term:

`(x-2)^2`

will be always a positive number, no matter what value takes x, it will be always positive because it is squared.

The other term is a positive number which is 5. So a random positive number plus 5 will be always another positive number. So the graph never takes negative values at y-axis, therefore it opens up.

Hence, this statement is wrong

The graph contains the point (2, 5).

As we saw in the first statement, the vertex of the function is (2,5). So according to that, we can conclude easily that the graph contains the point (2, 5).

Hence, this statement is true.

The graph intersects the x-axis at one unique point.

As we saw in the second statement, there are no real roots for the function. So, the graph never intersects the x-axis.

Hence, this statement is wrong

I attached a picture of the graph.