Question

Given the graph y=x^2+4x+3. How to find the vertex of the quadratic equation.

Answer 1

\(\bf \textit{vertex of a parabola}\\ \quad \\
y = {\color{red}{ a}}x^2+{\color{blue}{ b}}x+{\color{green}{ c}}\qquad

\qquad \quad
\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\)

Answer 2

But, how do I solve this problem?

Answer 3

I don't understand how.

Answer 4

you plug in your values \(\bf \textit{vertex of a parabola}\\ \quad \\
y = {\color{red}{ 1}}x^2{\color{blue}{ +4}}x{\color{green}{ +3}}\qquad

\qquad \quad
\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\)

Answer 5

So it would be 4/2 and 3-4/4?

Answer 6

Which would be (2,-3)?

Answer 7

close, but not there yet \(\bf \left(-\cfrac{{\color{blue}{ 4}}}{2{\color{red}{ (1)}}}\quad ,\quad {\color{green}{ 3}}-\cfrac{{\color{blue}{ 4}}^2}{4{\color{red}{ (1)}}}\right)\)