how do you find a vertex in intercept form

Sagot :

Quadratic functions can be written in vertex form, or

[tex]y=a\mleft(x-h\mright)^2+k[/tex]

This is especially useful because the vertex of the function is found at the point (h, k).

We can find this form by completing squares, for instance, let y be:

[tex]y=x^2+bx+c[/tex]

we can see that this equation is equal to

[tex]y=x^2+2(\frac{b}{2})x+(\frac{b}{2})^2-(\frac{b}{2})^2+c[/tex]

because

[tex]2(\frac{b}{2})=b[/tex]

and

[tex](\frac{b}{2})^2-(\frac{b}{2})^2=0[/tex]

However, in this form, we can see that the first 3 terms are a perfect square, that is

[tex]x^2+2(\frac{b}{2})x+(\frac{b}{2})^2=(x+\frac{b}{2})^2[/tex]

hence,

[tex]\begin{gathered} y=x^2+bx+c \\ y=(x+\frac{b}{2})^2-(\frac{b}{2})^2+c \end{gathered}[/tex]

If we define

[tex]\begin{gathered} -(\frac{b}{2})^2+c=k \\ \text{and} \\ h=\frac{b}{2} \end{gathered}[/tex]

we have that

[tex]y=(x+h)^2+k[/tex]

the constant a arise when you have a leading term different from 1 in x^2.