Question

Solving quadratics parabola
Y=x^2-4x+6

Answer 1

you want the x-interecepts?

Answer 2

*intercepts

Answer 3

This is factoring parabolas by the vertex and the zero

Answer 4

I don't know I just started parabolas an I'm confused

Answer 5

\(\large\color{black}{ \displaystyle y=x^2-4x+6 }\)
to find the x-intercepts, you need to know which value(s) to plug in for x, in order to get 0 (for y). This is why we set y=0, the following way:
\(\large\color{black}{ \displaystyle 0=x^2-4x+6 }\)

Now, we need to solve this quadratic equation. The fastest method I see is completing the square. (you can choose another methdo if you don't like this one)
\(\large\color{black}{ \displaystyle 0=x^2-4x+6 }\)
I will make the right hand side into a perfect square trinomial, by subtracting 2 from both sides.
\(\large\color{black}{ \displaystyle 0\color{red}{-2} =x^2-4x+6\color{red}{-2} }\)
\(\large\color{black}{ \displaystyle -2=x^2-4x+4 }\)
then, I will factor the right side
\(\large\color{black}{ \displaystyle -2=(x-2)(x-2) }\)
\(\large\color{black}{ \displaystyle -2=(x-2)^2 }\)

Answer 6

Now, all that remains is to:
\(\small\color{black}{ \bullet }\) Square root both sides.
\(\small\color{black}{ \bullet }\) Add 2 to both sides.

There are your solutions \(\small\color{black}{ {\bf!} }\)

Answer 7

They all are imaginary however. We can prove that:

\(\large\color{black}{ \displaystyle y=x^2-4x+6}\)
\(\large\color{black}{ \displaystyle y=x^2-4x+4+2}\)
\(\large\color{black}{ \displaystyle y=(x-2)^2+2}\)

\(\large\color{black}{ \displaystyle {\rm Vertex}=(2,2)}\)

Answer 8

Since vertex is above the x-axis, and this parabola is opening up (saying that opening up, implies that its Y well get only larger than the vertex but not smaller than the vertex)

AND THEREFORE, this parabola does not intersect the x-axis.