Question

write the sum and product of zeros of the polynomial :5x^2-9x-3 . ​

Answer 1

Answer:

$\mathrm{Domain\:of\:}\:5x^2-9x-3\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<x<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}$

$\mathrm{Range\:of\:}5x^2-9x-3:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:-\frac{141}{20}\:\\ \:\mathrm{Interval\:Notation:}&\:[-\frac{141}{20},\:\infty \:)\end{bmatrix}$

$\mathrm{Axis\:interception\:points\:of}\:5x^2-9x-3:$ $\mathrm{X\:Intercepts}:$ $\left(\frac{9+141^{\frac{1}{2}}}{10},\:0\right),\:\left(\frac{9-141^{\frac{1}{2}}}{10},\:0\right),$ $\:\mathrm{Y\:Intercepts}:\:\left(0,\:-3\right)$

$\mathrm{X\:Intercepts}:\:\left(\frac{9+141^{\frac{1}{2}}}{10},\:0\right),\:\left(\frac{9-141^{\frac{1}{2}}}{10},\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:-3\right)$

$\mathrm{Vertex\:of}\:5x^2-9x-3:\quad \mathrm{Minimum}\space\left(\frac{9}{10},\:-\frac{141}{20}\right)$