Answer
The axis of symmetry is $x=-\displaystyle \frac{3}{2}$.
domain: $(-\infty,\infty)$
range: $[-\displaystyle \frac{49}{4},\infty)$
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Work Step by Step
First, rewrite $f(x)=ax^{2}+bx+c$ in standard form, $f(x)=a(x-h)^{2}+k$
Graphing:
1. Determine whether the parabola opens upward or downward.
If $a>0$, it opens upward. If $a<0$, it opens downward.
2. Determine the vertex of the parabola. The vertex is $(h, k)$.
3. Find any x-intercepts by solving $f(x)=0$.
The function's real zeros are the x-intercepts.
4. Find the y-intercept by computing $f(0)$.
5. Plot the intercepts, the vertex, and additional points as necessary
Connect these points with a smooth curve that is shaped like a bowl or an inverted bowl.
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$f(x)=x^{2}+3x-10$
$f(x)=(x^{2}+2(\displaystyle \frac{3}{2})x+\frac{9}{4}-\frac{9}{4})-10$
$f(x)=(x+\displaystyle \frac{3}{2})^{2}-\frac{49}{4}$
1. opens up (a=$1>0$).
2. vertex: $(-\displaystyle \frac{3}{2},-\frac{49}{4}). $The axis of symmetry is $x=-\displaystyle \frac{3}{2}.$
3. x-intercepts:
$(x+\displaystyle \frac{3}{2})^{2}-\frac{49}{4}=0$
$(x+\displaystyle \frac{3}{2})^{2}=\frac{49}{4}$
$(x+\displaystyle \frac{3}{2})=\pm\frac{7}{2}$
$x=-\displaystyle \frac{3}{2}\pm\frac{7}{2}$
$x=2$ ,$\quad x=-5$
4. y-intercept:
$f(x)=0^{2}+3(0)-10=-10$
The axis of symmetry is $x=-\displaystyle \frac{3}{2}$.
domain: $(-\infty,\infty)$
range: $[-\displaystyle \frac{49}{4},\infty)$
