Answer:
[tex]A.\ y = x^2 - 6x + 13[/tex] is the correct answer.
Step-by-step explanation:
We know that vertex equation of a parabola is given as:
[tex]y = a(x-h)^2+k[/tex]
where [tex](h,k)[/tex] is the vertex of the parabola and
[tex](x,y)[/tex] are the coordinate of points on parabola.
As per the question statement:
The parabola opens upwards that means coefficient of [tex]x^{2}[/tex] is positive.
Let [tex]a = +1[/tex]
Minimum of parabola is at x = 3.
The vertex is at the minimum point of a parabola that opens upwards.
[tex]\therefore[/tex] [tex]h = 3[/tex]
Putting value of a and h in the equation:
[tex]y = 1(x-3)^2+k\\\Rightarrow y = (x-3)^2+k\\\Rightarrow y = x^2-6x+9+k[/tex]
Formula used: [tex](a-b)^2=a^{2} +b^{2} -2\times a \times b[/tex]
Comparing the equation formulated above with the options given we can observe that the equation formulated above is most similar to option A.
Comparing [tex]y = x^2 - 6x + 13[/tex] and [tex]y = x^2-6x+9+k[/tex]
13 = 9+k
k = 4
Please refer to the graph attached.
Hence, correct option is [tex]A.\ y = x^2 - 6x + 13[/tex]
Answer:
A. y = x^2 -6x + 13
Step-by-step explanation:
