Draw this graph [0;270] -1/2tan+3/4 And find the domain and the range and period

The graph is a transformation of the basic tangent function $y = \tan(x)$.

• The $-\frac{1}{2}$ factor vertically compresses the graph and reflects it across the x-axis.
• The $+\frac{3}{4}$ shifts the graph vertically upwards by $\frac{3}{4}$ units.

The period of $y = a \tan(bx + c) + d$ is $\frac{\pi}{|b|}$ (in radians) or $\frac{180^\circ}{|b|}$ (in degrees). In this function, $y = -\frac{1}{2}\tan(x) + \frac{3}{4}$, $b=1$.
So, the period is $\frac{180^\circ}{1} = 180^\circ$.

The basic tangent function $y = \tan(x)$ has vertical asymptotes at $x = 90^\circ + 180^\circ n$, where $n$ is an integer.
For $y = -\frac{1}{2}\tan(x) + \frac{3}{4}$, the vertical asymptotes remain at $x = 90^\circ + 180^\circ n$.
Within the interval $[0^\circ, 270^\circ]$, the asymptotes are at $x = 90^\circ$ and $x = 90^\circ + 180^\circ = 270^\circ$.

The horizontal shift line is $y = \frac{3}{4}$. We set $-\frac{1}{2}\tan(x) + \frac{3}{4} = \frac{3}{4}$.
$-\frac{1}{2}\tan(x) = 0$
$\tan(x) = 0$
This occurs at $x = 0^\circ, 180^\circ, 360^\circ, \dots$.
Within the interval $[0^\circ, 270^\circ]$, these points are $x = 0^\circ$ and $x = 180^\circ$.

• At $x = 0^\circ$: $y = -\frac{1}{2}\tan(0^\circ) + \frac{3}{4} = -\frac{1}{2}(0) + \frac{3}{4} = \frac{3}{4}$. Point: $(0^\circ, \frac{3}{4})$.
• At $x = 180^\circ$: $y = -\frac{1}{2}\tan(180^\circ) + \frac{3}{4} = -\frac{1}{2}(0) + \frac{3}{4} = \frac{3}{4}$. Point: $(180^\circ, \frac{3}{4})$.
• As $x$ approaches $90^\circ$ from the left ($x \to 90^{\circ-}$), $\tan(x) \to +\infty$, so $-\frac{1}{2}\tan(x) \to -\infty$, and $y \to -\infty$.
• As $x$ approaches $90^\circ$ from the right ($x \to 90^{\circ+}$), $\tan(x) \to -\infty$, so $-\frac{1}{2}\tan(x) \to +\infty$, and $y \to +\infty$.
• As $x$ approaches $270^\circ$ from the left ($x \to 270^{\circ-}$), $\tan(x) \to +\infty$, so $-\frac{1}{2}\tan(x) \to -\infty$, and $y \to -\infty$.

Plot the points $(0^\circ, \frac{3}{4})$ and $(180^\circ, \frac{3}{4})$. Draw vertical asymptotes at $x = 90^\circ$ and $x = 270^\circ$. Sketch the curve, remembering that the negative coefficient flips the usual upward slope of the tangent function. The graph will decrease from left to right between asymptotes.

(Imagine a graph here with vertical asymptotes at $x=90^\circ$ and $x=270^\circ$. The curve passes through $(0^\circ, 0.75)$ and $(180^\circ, 0.75)$. Between $0^\circ$ and $90^\circ$, the curve goes down from 0.75 towards $-\infty$. Between $90^\circ$ and $180^\circ$, the curve comes from $+\infty$ and goes down to 0.75. Between $180^\circ$ and $270^\circ$, the curve goes down from 0.75 towards $-\infty$.)

The domain is restricted by the vertical asymptotes. The function is undefined at $x = 90^\circ$ and $x = 270^\circ$.
The given interval is $[0^\circ, 270^\circ]$.
So, the domain is $[0^\circ, 90^\circ) \cup (90^\circ, 270^\circ)$.

The tangent function, even with transformations, covers all real numbers. The vertical shift and compression/reflection do not change this.
The range is $(-\infty, \infty)$.

The period is $180^\circ$.

The domain is $\boxed{[0^\circ, 90^\circ) \cup (90^\circ, 270^\circ]}$.
The range is $\boxed{(-\infty, \infty)}$.
The period is $\boxed{180^\circ}$.