Question

Solve this quadratic function

Answer 1

`\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-1}x^2\stackrel{\stackrel{b}{\downarrow }}{+3}x\stackrel{\stackrel{c}{\downarrow }}{+8} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{3}{2(-1)}~~,~~8-\cfrac{3^2}{4(-1)} \right)\implies \left( \cfrac{3}{2}~~,~~ 8+\cfrac{9}{4}\right)\implies \left( \cfrac{3}{2}~~,~~ \cfrac{41}{4}\right)`

now, for the solutions, or namely the x-intercepts, we simply set y = 0 and solve for "x", now this one doesn't factor into integers nicely, so we'll need to use the quadratic formula to get them.

`~~~~~~~~~~~~\textit{quadratic formula} \\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{-1}x^2\stackrel{\stackrel{b}{\downarrow }}{+3}x\stackrel{\stackrel{c}{\downarrow }}{+8} \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ x=\cfrac{ - 3 \pm \sqrt { 3^2 -4(-1)(8)}}{2(-1)}\implies x = \cfrac{-3\pm√(9+32)}{-2}\implies x = \cfrac{3\mp√(41)}{2}`

`x = \begin{cases} \cfrac{3-√(41)}{2}\\[2em] \cfrac{3+√(41)}{2} \end{cases}\qquad \leftarrow \qquad na mely\textit{ solutions or zeros or x-intercepts}`