Question

Which equation represents a parabola that opens upward, has a minimum at x = 3, and has a line of symmetry at x = 3?

Answer 1

Answer:

A. y = x^2 -6x + 13

Step-by-step explanation:

Answer 2

Answer:

$A.\ y = x^2 - 6x + 13$ is the correct answer.

Step-by-step explanation:

We know that vertex equation of a parabola is given as:

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

where $(h,k)$ is the vertex of the parabola and

$(x,y)$ are the coordinate of points on parabola.

As per the question statement:

The parabola opens upwards that means coefficient of $x^{2}$ is positive.

Let $a = +1$

Minimum of parabola is at x = 3.

The vertex is at the minimum point of a parabola that opens upwards.

$\therefore$ $h = 3$

Putting value of a and h in the equation:

$y = 1(x-3)^2+k\\\Rightarrow y = (x-3)^2+k\\\Rightarrow y = x^2-6x+9+k$

Formula used: $(a-b)^2=a^{2} +b^{2} -2\times a \times b$

Comparing the equation formulated above with the options given we can observe that the equation formulated above is most similar to option A.

Comparing $y = x^2 - 6x + 13$ and $y = x^2-6x+9+k$

13 = 9+k

k = 4

Please refer to the graph attached.

Hence, correct option is $A.\ y = x^2 - 6x + 13$