Question

Im having troubles with the last question, any help?

Answer 1

the x-intercepts are simply found by setting y = 0 and then solving for "x", now, this equation doesn't factor in nice integers, so let's plug it in the quadratic formula.

`\stackrel{f(x)}{0}~~ = ~~4x^2+10x-7 \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{4}x^2\stackrel{\stackrel{b}{\downarrow }}{+10}x\stackrel{\stackrel{c}{\downarrow }}{-7} \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ x= \cfrac{ - (10) \pm \sqrt { (10)^2 -4(4)(-7)}}{2(4)}\implies x=\cfrac{-10\pm√(100+112)}{8}`

`x=\cfrac{-10\pm√(212)}{8}\implies x=\cfrac{-10\pm 2√(53)}{8}\implies x=\cfrac{-5\pm √(53)}{4} \\\\\\ x= \begin{cases} (-5 + √(53))/(4)\\\\ (-5 - √(53))/(4) \end{cases}\qquad \qquad x\approx \begin{cases} 0.57\\ -3.07 \end{cases}`