Quadratic graphs and extrema

Every quadratic $y = ax^2 + bx + c$ draws the same family of curve — a parabola — and yet the three numbers $a$ , $b$ , $c$ pin down everything about it before you plot a single point: which way it opens, exactly where it turns, whether it touches the x-axis once, twice, or never. That predictive power comes from one piece of algebra: completing the square. Rewriting $ax^2+bx+c$ in vertex form exposes the turning point directly, without any calculus, and the same rewrite reveals a clean formula for the maximum or minimum value the expression can ever take. This article extends the Class 11 Algebra picture of quadratics — building on the discriminant you already met when finding roots — into a full graphical toolkit: vertex, axis of symmetry, intercepts, and extrema, all read off algebraically from $a$ , $b$ , $c$ .

The core insight

Completing the square on $y=ax^2+bx+c$ gives the vertex $\left(-\dfrac{b}{2a},\,-\dfrac{D}{4a}\right)$ , so the sign of $a$ decides maximum versus minimum and the sign of $D=b^2-4ac$ decides how many times the parabola meets the x-axis.

The core idea

Vertex form comes from completing the square. Starting from $y=ax^2+bx+c$ , factor out $a$ from the quadratic and linear terms, then add and subtract the square needed to complete a perfect square:

y = a\left(x+\frac{b}{2a}\right)^2 + \left(c-\frac{b^2}{4a}\right) = a\left(x+\frac{b}{2a}\right)^2 - \frac{D}{4a},

where: $D=b^2-4ac$ is the discriminant, and the bracketed square term is always $\ge 0$ (or $\le 0$ if $a<0$ flips the sign outside). Because $\left(x+\frac{b}{2a}\right)^2$ is smallest — exactly zero — when $x=-\frac{b}{2a}$ , this single rewrite locates the turning point without any graph paper.

Reading off the vertex. The turning point (vertex) of the parabola sits at

x = -\frac{b}{2a}, \qquad y = -\frac{D}{4a} = \frac{4ac-b^2}{4a}.

where: the x-coordinate is the axis of symmetry, and the y-coordinate is the extreme value the expression attains there.

Shape and extrema. If $a>0$ the parabola opens upward and the vertex is a minimum; if $a<0$ it opens downward and the vertex is a maximum. There is never both — a parabola has exactly one extremum, never the other kind, because it keeps rising (or falling) without bound on either side of the vertex.

x-intercepts and the discriminant. Setting $y=0$ gives $ax^2+bx+c=0$ , whose roots are the x-intercepts. So $D>0$ means two crossing points, $D=0$ means the parabola just touches the axis at the vertex itself, and $D<0$ means the curve never meets the x-axis — it sits entirely above (if $a>0$ ) or entirely below (if $a<0$ ) it. The y-intercept is always the point $(0,c)$ , read straight off by setting $x=0$ . The vertical line $x=-b/(2a)$ through the vertex is the axis of symmetry: the parabola is a mirror image of itself across this line, so the two x-intercepts (when they exist) are always equidistant from it — in fact their midpoint is $-b/(2a)$ .

Connecting back to the discriminant. Because the vertex y-coordinate equals $-D/(4a)$ , the sign of $D$ tells you exactly where the vertex sits relative to the x-axis, and that in turn fixes how many times the curve crosses it. For an upward parabola ( $a>0$ ): if $D<0$ the vertex y-value is positive, so the whole curve sits above the axis and there are no real roots; if $D=0$ the vertex sits exactly on the axis, giving one repeated root; if $D>0$ the vertex dips below the axis, so the curve must cross it twice on the way up on either side. The same reasoning, mirrored, applies to a downward parabola. This is the graphical picture behind the "nature of roots" rule you already used with the discriminant — the algebra and the graph are telling the same story from two angles.

Figure

Two parabolas side by side, one opening upward with a minimum vertex and one opening downward with a maximum vertex, each with its axis of symmetry marked

Derivation

Derivation ALocating the vertex by completing the square

The goal is to show, starting only from the general quadratic $y=ax^2+bx+c$ and the algebraic technique of completing the square, that the turning point always sits at $x=-b/(2a)$ with extreme value $y=-D/(4a)$ , so that the vertex of any parabola can be read off directly from $a$ , $b$ , $c$ without plotting a single point.

Step 1. Factor $a$ out of the first two terms only: $y = a\left(x^2+\dfrac{b}{a}x\right)+c$ . This isolates the part that needs completing, leaving $c$ untouched outside the bracket.

Step 2. Complete the square inside the bracket. Half of $b/a$ is $b/(2a)$ , and its square is $b^2/(4a^2)$ ; add and subtract this inside the bracket: $y = a\left(x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}-\dfrac{b^2}{4a^2}\right)+c = a\left(x+\dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a}+c.$

Step 3. Combine the constant terms over $D=b^2-4ac$ . Write $c-\dfrac{b^2}{4a} = \dfrac{4ac-b^2}{4a} = -\dfrac{b^2-4ac}{4a} = -\dfrac{D}{4a}$ , giving the vertex form $y=a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{D}{4a}$ .

Step 4. Read off the vertex. Since $\left(x+\dfrac{b}{2a}\right)^2 \ge 0$ for every real $x$ , and equals zero exactly when $x=-\dfrac{b}{2a}$ , the bracketed term contributes its smallest possible size at that x-value. If $a>0$ , multiplying a non-negative square by a positive $a$ keeps it non-negative, so $y$ is smallest there — a minimum of $-D/(4a)$ ; if $a<0$ , the same square multiplied by a negative $a$ is largest (least negative) there, giving a maximum of $-D/(4a)$ . Either way the vertex sits at $\left(-\dfrac{b}{2a},\,-\dfrac{D}{4a}\right)$ , exactly as claimed.

Worked example

Sketch $y=x^2-4x+3$ fully — vertex, axis of symmetry, intercepts, and state its extremum — then use the same method on $y=-2x^2+8x-5$ to find its maximum.

Step 1. Identify $a,b,c$ for the first quadratic: $a=1$ , $b=-4$ , $c=3$ . Since $a>0$ , the parabola opens upward and its vertex is a minimum.

Step 2. Find the vertex using $x=-b/(2a)$ : $x=-\dfrac{-4}{2(1)}=2$ , and $y=(2)^2-4(2)+3=4-8+3=-1$ . So the vertex is $(2,-1)$ , the minimum value of $y$ is $-1$ , and the axis of symmetry is the line $x=2$ .

Step 3. Find the intercepts. Solving $x^2-4x+3=0$ factors as $(x-1)(x-3)=0$ , giving x-intercepts $(1,0)$ and $(3,0)$ — note their midpoint is $2$ , matching the axis of symmetry. Setting $x=0$ gives the y-intercept $(0,3)$ .

Step 4. Cross-check via completing the square directly: $y=x^2-4x+3=(x-2)^2-4+3=(x-2)^2-1$ , confirming the vertex $(2,-1)$ since $(x-2)^2\ge0$ with equality only at $x=2$ . The parabola opens upward through $(1,0)$ and $(3,0)$ , dips to its minimum at $(2,-1)$ , and crosses the y-axis at $(0,3)$ .

Competitive stretch

Now find the maximum value of $y=-2x^2+8x-5$ . Here $a=-2$ , $b=8$ , $c=-5$ , and $a<0$ , so the vertex is a maximum. The x-coordinate is $x=-\dfrac{8}{2(-2)}=2$ , and substituting gives $y=-2(2)^2+8(2)-5=-8+16-5=3$ . Checking against $-D/(4a)$ : $D=8^2-4(-2)(-5)=64-40=24$ , so $-D/(4a)=-24/(-8)=3$ ✓. The maximum value of $y$ is $3$ , occurring at $x=2$ .

Common traps

Forgetting the negative sign — the vertex x-coordinate is $x=-b/(2a)$ , not $x=b/(2a)$ ; this single sign slip throws off every other feature of the sketch.
Assuming every parabola has a minimum. Whether the vertex is a maximum or minimum depends entirely on the sign of $a$ ; a downward-opening parabola has no minimum at all.
Mixing up this algebraic completing-the-square method with the Class XII calculus method of setting $dy/dx=0$ . Only the algebraic route is examinable here, though both agree on the same point.
Reading the vertex y-coordinate as $+D/(4a)$ instead of $-D/(4a)$ — carry the minus sign through from the vertex-form derivation carefully.
Sketching the wrong orientation before checking the sign of $a$ — always decide upward versus downward first, since every other feature (which way the vertex "type" goes, where the curve rises or falls) depends on it.
Assuming a downward parabola with $D<0$ has no graph at all — it still exists and still has a vertex and a y-intercept; it simply never touches the x-axis, sitting entirely below it.

Ask the tutor: If completing the square still feels shaky, ask the AI tutor to walk through the four algebra steps on your own quadratic, one line at a time.

Why does completing the square give the vertex, but factoring gives the roots?

They answer different questions from the same quadratic. Factoring $ax^2+bx+c$ finds the x-values where $y=0$ — the roots or x-intercepts. Completing the square instead rewrites the whole expression as $a(\text{square})+\text{constant}$ , which exposes the single x-value where the square term vanishes and $y$ hits its extreme value. Both techniques operate on the same $a,b,c$ , but factoring targets $y=0$ specifically while completing the square targets the turning point for any $y$ .

Does the vertex formula work even if the quadratic has no real roots?

Yes — the vertex formula $x=-b/(2a)$ , $y=-D/(4a)$ uses only $a$ , $b$ , $c$ and never requires the roots to exist. If $D<0$ the parabola simply never touches the x-axis, but it still has a well-defined vertex and axis of symmetry; you can sketch the whole curve — shape, turning point, y-intercept — without ever solving $ax^2+bx+c=0$ .

How is this different from the maxima and minima I will study with derivatives later?

The underlying point on the graph is identical — same vertex, same extreme value — but the method here is purely algebraic: complete the square and read off the coordinates. In Class XII you will instead set $dy/dx=2ax+b=0$ , solve for $x=-b/2a$ , and confirm it is a maximum or minimum using the second derivative's sign. Both routes land on the same answer for a quadratic; completing the square is simply the school-algebra version that does not need calculus.