Answer:
[tex]5\left(x+\dfrac{9}{10}\right)^2-\dfrac{1}{20}[/tex]
Step-by-step explanation:
Given:
[tex]5x^2+9x+4[/tex]
Factor out the coefficient of the leading term from the first two terms:
[tex]\implies 5\left(x^2+\dfrac{9}{5}x\right)+4[/tex]
Add the square of half the coefficient of the x term inside the parentheses and subtract its distributed value from the expression:
[tex]\implies 5\left(x^2+\dfrac{9}{5}x+\left(\dfrac{\frac{9}{5}}2\right)^2\right)+4-5\left(\dfrac{\frac{9}{5}}2\right)^2[/tex]
Simplify:
[tex]\implies 5\left(x^2+\dfrac{9}{5}x+\left(\dfrac{9}{10}\right)^2\right)+4-5\left(\dfrac{9}{10}\right)^2[/tex]
[tex]\implies 5\left(x^2+\dfrac{9}{5}x+\dfrac{81}{100}\right)+4-5\left(\dfrac{81}{100}\right)[/tex]
[tex]\implies 5\left(x^2+\dfrac{9}{5}x+\dfrac{81}{100}\right)+4-\dfrac{405}{100}[/tex]
[tex]\implies 5\left(x^2+\dfrac{9}{5}x+\dfrac{81}{100}\right)-\dfrac{1}{20}[/tex]
Factor the perfect trinomial contained within the parentheses:
[tex]\implies 5\left(x+\dfrac{9}{10}\right)^2-\dfrac{1}{20}[/tex]
Therefore, the given expression written in vertex form is:
[tex]5\left(x+\dfrac{9}{10}\right)^2-\dfrac{1}{20}[/tex]
