how do I find the vertex and intercepts of 1/3(x^2 +5x-4)?

Question

muzzack · Answer

Final result :

  $\huge\frac{x^2 + 5x - 4}{3}$

zzr0ck3r · Answer

$f(x) = \frac{1}{3}(x^2+5x-4)=\frac{1}{3}(x+\frac{5}{2})^2-4-\frac{1}{3}(\frac{5}{2})^2$ Simplify this and you have vertex form.

triciaal · Answer

vertex x value is -b/2a and y value is f(x) from the eqn in standard form

anonymous · Answer

$\Large \frac{1}{3}(x^2+5x-4)=\frac{x^2}{3} + \frac{5}{3}x-\frac{4}{3}$

The given equation is $\Large y = \frac{x^2}{3} + \frac{5}{3}x-\frac{4}{3}$
This is of the form $\Large ax^2 + bx + c = y $
The $\Large x$ co-ordinate of the vertex $\Large = \frac{-b}{2a}=\huge \frac{\frac{-5}{3}}{\frac{2}{3}}=\frac{-5}{2}$
Replacing $\Large x$ by $\Large \frac{-5}{2}$ in the given equation, 
$\Large y=\frac{1}{3}\times (\frac{-5}{2})^2+\frac{5}{3}\times \frac{-5}{2}-\frac{4}{3}=\frac{-41}{12}$

The vertex =$\huge \color{red}{ (\frac{-5}{2},\frac{-41}{12}) }$


To find the y intercept we replace x by 0. 
$\Large \Rightarrow y=\frac{-4}{3}$
To find the x intercept we set y = 0.
$\Large \Rightarrow \frac{x^2}{3} + \frac{5}{3}x-\frac{4}{3}=0$
$\huge \Rightarrow x=\frac{-5\pm \sqrt{41}}{2}$

Thus the $\Large x$ intercepts are $\huge \color{red}{\frac{-5-\sqrt{41}}{2}}$ and $\huge \color{red}{\frac{-5+\sqrt{41}}{2}}$ while the $\Large y$ intercept is $\huge \color{red}{\frac{-4}{3}}$