Q:

Y=(x+5) (x+4) in vertex form of parabola

Accepted Solution

A:
Answer:Step-by-step explanation:The vertex form of an equation of a parabola y = ax² + bx + c:[tex]y=a(x-h)^2+k[/tex](h, k) - verex[tex]h=\dfrac{-b}{2a},\ k=f(h)[/tex]We have:[tex]y=(x+5)(x+4)[/tex]            use FOIL (a + b)(c + d) = ac + ad + bc + bd[tex]y=(x)(x)+(x)(4)+(5)(x)+(5)(4)=x^2+4x+5x+20=x^2+9x+20[/tex]a = 1, b = 9, c = 20[tex]h=\dfrac{-9}{2(1)}=\dfrac{-9}{2}=-\dfrac{9}{2}[/tex]For k, put the value of h to the equation y = (x + 5)(x + 4):[tex]k=\left(-\dfrac{9}{2}+5\right)\left(-\dfrac{9}{2}+4\right)\\\\k=\left(-\dfrac{9}{2}+\dfrac{10}{2}\right)\left(-\dfrac{9}{2}+\dfrac{8}{2}\right)\\\\k=\left(\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)=-\dfrac{1}{4}[/tex]Finally:[tex]y=1\bigg(x-\left(-\dfrac{9}{2}\right)\bigg)^2-\dfrac{1}{4}\\\\y=\left(x+\dfrac{9}{2}\right)^2-\dfrac{1}{4}[/tex]