Question

Find the zeros of y=3x^2+x-2, then use the zeros to find the vertex of the graph of the parabola.

Answer 1

first off, let's find the zeros, keeping in mind that, the vertex of the parabola, will be half-way between the zeros.


`\bf y=3x^2+x-2\implies 0=3x^2+x-2\implies 0=(3x-2)(x+1) \\\\\\ 0=3x-2\implies 2=3x\implies \boxed{\cfrac{2}{3}=x} \\\\\\ and \\\\\\ 0=x+1\implies \boxed{-1=x}`

now, the zeros are there, now, what's half-way? or the midpoint of that interval? well, from -1 to 0, is say 3/3, from 0 to 2/3 is 2/3, we add both to get 3/3 + 2/3 the interval is 5/3.

now, what's half of 5/3?
`\bf \cfrac{\quad(5)/(3) \quad }{2}\implies \cfrac{\quad(5)/(3) \quad }{(2)/(1)}\implies \cfrac{5}{3}\cdot \cfrac{1}{2}\implies \cfrac{5}{6}`

now, 5/3 is the same as 10/6, so if we move from say the -1 spot over 5/6 towards the 0, we'll end up at -1/6.

therefore half of 5/3 is 5/6, and if we move from either "zero" to the middle by 5/6 units, we'll end up at -1/6, check the picture below.

therefore, the vertex is right there, above -1/6, when x = -1/6, what is that anyway?


`\bf y=3x^2+x-2\qquad \boxed{x=-(1)/(6)}\qquad y=3\left( -(1)/(6) \right)^2+\left( -(1)/(6) \right)-2 \\\\\\ y=3\left( -(1)/(6) \right)\left( -(1)/(6) \right)-(1)/(6)+2\implies y=3\left( (1)/(36) \right)-(1)/(6)-2 \\\\\\ y=\cfrac{1}{12}-\cfrac{1}{6}-2\impliedby \textit{our LCD is 12}\implies y=\cfrac{1-2-24}{12} \\\\\\ \boxed{y=\cfrac{-25}{12}}`