finding domain and range of f(x) = -x² - 10x - 24

TL;DR:​ The domain of $f(x)=-x^2-10x-24$ is all real numbers: $(-\infty ,+\infty )$. The range is $(-\infty ,1]$ since the vertex is at $(-5,1)$ and the parabola opens downward.

Question

How do I find the domain and range for the quadratic equation: f(x) = -x² - 10x - 24?​

---

Answer

The interactive parabola is ready — open it and drag the k slider (or press Load f(x) = −x² − 10x − 24) to watch the shaded range band slide with the vertex while the domain refuses to budge.

Solving it step by step

Step 1 — Domain: can any $x$ break the rule?​

The expression $-x^2-10x-24$ is a polynomial. You can square any real number, multiply it by $-1$, and add real numbers — nothing ever goes undefined (no division, no square root, no logarithm). So every real number is allowed:

Step 2 — Rewrite in vertex form by completing the square.​

So $h=-5$, $k=1$: the vertex is $(-5,1)$, and since $a=-1<0$ the parabola opens downward.

Step 3 — Range: which outputs are actually reachable?​

This is where the vertex form earns its keep. Because $(x+5{)}^2\ge 0$ for every real $x$,

Equality holds exactly when $x=-5$, so $1$ is attained and it is the maximum. As $x$ runs off toward $\pm \infty$, $(x+5{)}^2$ grows without bound and $f(x)$ drops toward $-\infty$. Every value below $1$ is hit somewhere by the continuous curve, so:

Quick check with the vertex point: $f(-5)=-25+50-24=1$. ✓

• Key concepts

  • Domain of a polynomial — every real $x$ is admissible, so it is always $(-\infty ,+\infty )$; the domain question is only interesting when there is a division, root, or log.
  • Vertex form $y=a(x-h{)}^2+k$ — completing the square converts $a{x}^2+bx+c$ into this shape; the vertex is $(h,k)$.
  • Range of a quadratic — determined completely by the sign of $a$ and the vertex height $k$. $a>0$: $[k,+\infty )$; $a<0$: $(-\infty ,k]$.
  • Boundedness — $(x-h{)}^2\ge 0$ is the whole reason the quadratic is bounded on one side; this is the algebraic engine of the range.

• Confusion points

  • Vertex $x$-coordinate vs range bound — the range is governed by $k$ (the $y$-value, here $1$), not by $h=-5$. Students often quote $-5$ as the range because it is the "important number" they just computed.
  • Sign flip when reading $h$​ — $f(x)=-(x+5{)}^2+1$ shifts left by $5$, so $h=-5$, not $+5$. The bracket shows $+5$, the vertex coordinate is $-5$.
  • Domain vs range — the domain is the set of inputs ($x$, horizontal), the range is the set of outputs ($y$, vertical). The teal band in the tool is vertical extent — that is the range.
  • Open vs closed endpoint — the maximum $1$ is achieved (at $x=-5$), so the bracket at $1$ is closed: $(-\infty ,1]$, never $(-\infty ,1)$.
  • "Opens up means range is positive" — the sign of $a$ fixes the direction of the unbounded end, not whether outputs are positive. Here the outputs are mostly negative; the ceiling is just $1$.

Understanding check: A classmate solves $g(x)=2(x-4{)}^2-7$ and writes "range $=[4,+\infty )$ because the vertex is at $(4,-7)$." Which number did they misuse, and what is the correct range?