Which equation represents a parabola that opens upward, has a minimum at x = 3, and has a line of symmetry at x = 3? A. y = x^2 - 6x + 13 B. y = x^2 - 8x + 19 C. y= x^2 - 3x + 6…

Question

asked Feb 22, 2021 142k views

2 Answers

Pavel Donchev · Answer 1 · 2021-02-25T16:02:10+0000

Answer:

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

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* a * 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
$A.\ y = x^2 - 6x + 13$