Question #f37a1

May 13, 2017

Vertex is at $\left(3.5 , - 0.5\right)$ , axis of symmetry is $x = 3.5$,
y-intercept is at $\left(0 , 24\right)$, x-intercepts are at $\left(3 , 0\right) \mathmr{and} \left(4 , 0\right)$

Explanation:

$F \left(x\right) = 2 {x}^{2} - 14 x + 24 = 2 \left({x}^{2} - 7 x + {\left(\frac{7}{2}\right)}^{2}\right) - \frac{49}{2} + 24 \mathmr{and} F \left(x\right) = 2 {\left(x - \frac{7}{2}\right)}^{2} - \frac{1}{2}$
Comparing with standard equation $F \left(x\right) = a {\left(x - h\right)}^{2} + f , \left(h , f\right)$ being vertex , we get vertex as
$h = \frac{7}{2} , f = - \frac{1}{2} \mathmr{and} \left(3.5 , - 0.5\right)$ Axis of symmetry is $x = 3.5$

y-intercept can be obtained by putting $x = 0$ in the equation
$\therefore F \left(x\right) \mathmr{and} y = 24$
y - intercept is at $\left(0 , 24\right)$

x-intercept can be obtained by putting $F \left(x\right) \mathmr{and} y = 0$ in the equation

$2 {\left(x - \frac{7}{2}\right)}^{2} - \frac{1}{2} = 0 \mathmr{and} 2 {\left(x - \frac{7}{2}\right)}^{2} = \frac{1}{2} \mathmr{and} {\left(x - \frac{7}{2}\right)}^{2} = \frac{1}{4}$ or
${\left(x - \frac{7}{2}\right)}^{2} = \pm \sqrt{\frac{1}{4}} \therefore x = \frac{7}{2} + \frac{1}{2} = 4 \mathmr{and} x = \frac{7}{2} - \frac{1}{2} = 3$
x- intercepts are at $\left(3 , 0\right) \mathmr{and} \left(4 , 0\right)$ graph{2x^2-14x+24 [-40, 40, -20, 20]} [Ans]