# How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of y = [x^2 + 21x + 22]/16?

Feb 17, 2017

Vertex is $\left(- \frac{21}{2} , - \frac{353}{64}\right)$ and axis of symmetry is $x + \frac{353}{64} = 0$
$y$-intercept is $\frac{11}{8}$ and $x$-intercepts are $- 1.1$ and $- 19.9$.
Domain is $\left(- \infty , \infty\right)$, range is $\left[- 353 , 64 , \infty\right)$

#### Explanation:

The vertex form of quadratic equation is $y = a {\left(x - h\right)}^{2} + k$ and its vertex is $\left(h , k\right)$ and axis of symmetry is $x - h = 0$.

Hence we should first convert $y = \frac{{x}^{2} + 21 x + 22}{16}$ into vertex form.

$y = \frac{{x}^{2} + 21 x + 22}{16}$

= $\frac{1}{16} \left({x}^{2} + 2 \times \frac{21}{2} \times x + {\left(\frac{21}{2}\right)}^{2} - {\left(\frac{21}{2}\right)}^{2} + 22\right)$

= $\frac{1}{16} {\left(x + \frac{21}{2}\right)}^{2} - \frac{441}{64} + \frac{22}{16}$

= $\frac{1}{16} {\left(x + \frac{21}{2}\right)}^{2} - \frac{441 - 88}{64}$

= $\frac{1}{16} {\left(x - \left(- \frac{21}{2}\right)\right)}^{2} - \frac{353}{64}$

Therefore, its vertex is $\left(- \frac{21}{2} , - \frac{353}{64}\right)$ and axis of symmetry is $x - \left(- \frac{353}{64}\right) = 0$ or $x + \frac{353}{64} = 0$.

$y$-intercept can be found by putting $x = 0$ and $x$-intercept can be found by putting $y = 0$.

Hence $y$-intercept is $\frac{22}{16} = \frac{11}{8}$

and $x$-intercept are given by using quadratic formula on ${x}^{2} + 21 x + 22 = 0$ i.e. $x = \frac{- 21 \pm \sqrt{{21}^{2} - 4 \times 22}}{2} = \frac{- 21 \pm \sqrt{353}}{2}$ i.e. $- 1.1$ and $- 19.9$.

While domain is $\left(- \infty , \infty\right)$, range is $\left[- 353 , 64 , \infty\right)$
graph{(y-(x^2+21x+22)/16)=0 [-27.33, 12.67, -10.16, 9.84]}