10.5 Graphing quadratic equations  (Page 8/15)

In the previous set of exercises, you worked with the quadratic equation $R=\text{−}{x}^2+100x$ that modeled the revenue received from selling backpacks at a price of $x$ dollars. You found the selling price that would give the maximum revenue and calculated the maximum revenue. Now you will look at more characteristics of this model.
ⓐ Graph the equation $R=\text{−}{x}^2+100x$ . ⓑ Find the values of the x -intercepts.

For the revenue model in [link] and [link] , explain what the x -intercepts mean to the computer store owner.

For the revenue model in [link] and [link] , explain what the x -intercepts mean to the backpack retailer.

$x^2=100$

$y^2=144$

$m^2-40=0$

$n^2-80=0$

$4a^2=100$

$2b^2=72$

$r^2+32=0$

$t^2+18=0$

$\frac{4}{3}{v}^2+4=28$

$\frac{2}{3}{w}^2-20=30$

$5c^2+3=19$

$3d^2-6=43$

${\left(p-5\right)}^2+3=19$

${\left(q+4\right)}^2=9$

${\left(u+1\right)}^2=45$

${\left(z-5\right)}^2=50$

${\left(x-\frac{1}{4}\right)}^2=\frac{3}{16}$

${\left(y-\frac{2}{3}\right)}^2=\frac{2}{9}$

${\left(m-7\right)}^2+6=30$

${\left(n-4\right)}^2-50=150$

${\left(5c+3\right)}^2=\mathrm{-20}$

${\left(4c-1\right)}^2=\mathrm{-18}$

$m^2-6m+9=48$

$n^2+10n+25=12$

$64a^2+48a+9=81$

$4b^2-28b+49=25$

$x^2+22x$

$y^2+6y$

$m^2-8m$

$n^2-10n$

$a^2-3a$

$b^2+13b$

$p^2+\frac{4}{5}p$

$q^2-\frac{1}{3}q$

$c^2+20c=21$

$d^2+14d=\mathrm{-13}$

$x^2-4x=32$

$y^2-16y=36$

$r^2+6r=\mathrm{-100}$

$t^2-12t=\mathrm{-40}$

$v^2-14v=\mathrm{-31}$

$w^2-20w=100$

$m^2+10m-4=\mathrm{-13}$

$n^2-6n+11=34$

$a^2=3a+8$

$b^2=11b-5$

$\left(u+8\right)\left(u+4\right)=14$

$\left(z-10\right)\left(z+2\right)=28$

$3p^2-18p+15=15$

$5q^2+70q+20=0$

$4y^2-6y=4$

$2x^2+2x=4$

$3c^2+2c=9$

$4d^2-2d=8$

$4x^2-5x+1=0$

$7y^2+4y-3=0$

$r^2-r-42=0$

$t^2+13t+22=0$

$4v^2+v-5=0$

$2w^2+9w+2=0$

$3m^2+8m+2=0$

$5n^2+2n-1=0$

$6a^2-5a+2=0$

$4b^2-b+8=0$

$u\left(u-10\right)+3=0$

$5z(z-2)=3$

$\frac{1}{8}{p}^2-\frac{1}{5}p=-\frac{1}{20}$

$\frac{2}{5}{q}^2+\frac{3}{10}q=\frac{1}{10}$

$4c^2+4c+1=0$

$9d^2-12d=\mathrm{-4}$

1. ⓐ $9x^2-6x+1=0$
2. ⓑ $3y^2-8y+1=0$
3. ⓒ $7m^2+12m+4=0$
4. ⓓ $5n^2-n+1=0$

1. ⓐ $5x^2-7x-8=0$
2. ⓑ $7x^2-10x+5=0$
3. ⓒ $25x^2-90x+81=0$
4. ⓓ $15x^2-8x+4=0$

1. ⓐ $16r^2-8r+1=0$
2. ⓑ $5t^2-8t+3=9$
3. ⓒ $3{\left(c+2\right)}^2=15$

1. ⓐ $4d^2+10d-5=21$
2. ⓑ $25x^2-60x+36=0$
3. ⓒ $6{(5v-7)}^2=150$

Find two consecutive odd numbers whose product is 323.