ParabolaGraph of a Quadratic FunctionThe graph of a quadratic function $f(x) = ax^2 + bx + c$ (where $a \neq 0$) is a curve called a parabola.The direction of the branches depends on the sign of the leading coefficient $a$:

If $a > 0$, the branches point upward — the parabola has a lowest point, so the function has a minimum value.

If $a < 0$, the branches point downward — the parabola has a highest point, so the function has a maximum value.

For example, $y = 2x^2 + 3x - 2$ opens upward ($a = 2 > 0$), while $f(x) = -x^2 + 5x - 6$ opens downward ($a = -1 < 0$).The parabola is symmetric about a vertical line called the axis of symmetry, which passes through its turning point — the vertex.Vertex of a ParabolaThe vertex $T(r, k)$ of the parabola $f(x) = ax^2 + bx + c$ is its turning point (in Turkish exams it is called *tepe noktası*). Its coordinates are:$$r = -\frac{b}{2a}, \qquad k = f(r) = \frac{4ac - b^2}{4a}$$In practice, the easiest way to find $k$ is simply to substitute $x = r$ into the function: $k = f(r)$.Example: for $f(x) = x^2 + 2x + 1$ with vertex $T(r, k)$:$$r = -\frac{2}{2 \cdot 1} = -1, \qquad k = f(-1) = 1 - 2 + 1 = 0$$So $T(-1, 0)$ and the product $r \cdot k = (-1) \cdot 0 = 0$.The axis of symmetry of the parabola is the vertical line $x = r$.Minimum and Maximum of a Quadratic FunctionThe extreme value of $f(x) = ax^2 + bx + c$ is always attained at the vertex, i.e. at $x = r = -\frac{b}{2a}$:$$f_{\text{min/max}} = f\left(-\frac{b}{2a}\right) = \frac{4ac - b^2}{4a}$$

If $a > 0$: the function has a minimum equal to $k$, and no maximum.

If $a < 0$: the function has a maximum equal to $k$, and no minimum.

Quick algorithm:
1. Compute $r = -\frac{b}{2a}$.
2. Compute $k = f(r)$ — this is the minimum (if $a > 0$) or the maximum (if $a < 0$).For $y = 2x^2 + 3x - 2$: $r = -\frac{3}{4}$, so the minimum is $y\left(-\frac{3}{4}\right) = 2 \cdot \frac{9}{16} - \frac{9}{4} - 2 = -\frac{25}{8}$.Factored Form and RootsIf a quadratic function has roots $x_1$ and $x_2$ (that is, $f(x_1) = f(x_2) = 0$), it can be written in factored form:$$f(x) = a(x - x_1)(x - x_2)$$Key facts:

The $x$-coordinate of the vertex lies exactly midway between the roots: $r = \frac{x_1 + x_2}{2}$.

By Vieta's formulas for $f(x) = ax^2 + bx + c$: $x_1 + x_2 = -\frac{b}{a}$ and $x_1 \cdot x_2 = \frac{c}{a}$.

Typical problem: if $f(x) = x^2 + bx + c$ and $f(1) = f(2) = 0$, then $f(x) = (x-1)(x-2) = x^2 - 3x + 2$, so $b = -3$, $c = 2$ and $b + c = -1$.Finding a Parameter from the Vertex or ExtremumMany YÖS problems give the vertex or the max/min value and ask for an unknown parameter. Two standard schemes:1. The vertex abscissa is given. If the vertex of $f(x) = ax^2 + bx + c$ is at $x = r_0$, use the equation$$-\frac{b}{2a} = r_0$$For $f(x) = mx^2 - (2m-8)x + 1$ with vertex at $x = -1$: $\frac{2m-8}{2m} = -1 \Rightarrow 2m - 8 = -2m \Rightarrow m = 2$.2. The max/min value is given. If $\max f = k_0$ (or $\min f = k_0$), first find $r = -\frac{b}{2a}$, then solve$$f(r) = k_0$$If the vertex is given as $T(r_0, k_0)$, e.g. $T(1, t)$ or $T(2, k)$, substitute: the unknown ordinate is simply $k = f(r_0)$ after the parameter is found from $-\frac{b}{2a} = r_0$.Maximum of a Product of Linear ExpressionsIf $A$ and $B$ are linear expressions in $x$, their product $A \cdot B$ is a quadratic function, i.e. a parabola. To find $\max(A \cdot B)$ or $\min(A \cdot B)$:

Expand or find the roots of the product (each linear factor gives one root).

The extremum is at the midpoint of the roots: $r = \frac{x_1 + x_2}{2}$.

Substitute $x = r$ into $A \cdot B$.

Useful shortcut: if the sum $A + B$ is constant, the product $A \cdot B$ is maximal when $A = B$ (this happens exactly at the midpoint of the roots).For $A = 2x - 12$ and $B = -2x + 24$: $A + B = 12$ is constant, so the maximum is at $A = B = 6$, giving $\max(A \cdot B) = 36$.Parabola and a Line: Intersection and TangencyTo study the mutual position of the parabola $y = ax^2 + bx + c$ and the line $y = mx + n$, set them equal:$$ax^2 + bx + c = mx + n \;\Rightarrow\; ax^2 + (b - m)x + (c - n) = 0$$Let $D$ be the discriminant of this quadratic equation:

$D > 0$ — the line intersects the parabola at two points;

$D = 0$ — the line is tangent to the parabola (exactly one common point);

$D < 0$ — no common points.

Tangency condition: $D = 0$. This is the key equation for problems like "the line $y = 3x + 1$ is tangent to $y = x^2 - kx + 10$; find $k$" — it usually gives a quadratic equation in the parameter, and the answer often asks for the sum of its solutions (use Vieta: sum of values of $k$ equals $-\frac{b}{a}$ of that equation).Intersection Points with the Coordinate AxesFor the parabola $y = ax^2 + bx + c$:

Intersection with the $y$-axis: set $x = 0$, giving the point $(0, c)$. There is always exactly one such point.

Intersection with the $x$-axis: set $y = 0$ and solve $ax^2 + bx + c = 0$. The number of intersection points is determined by the discriminant $D = b^2 - 4ac$:

$D > 0$ — two points $(x_1, 0)$ and $(x_2, 0)$;

$D = 0$ — one point: the parabola touches the $x$-axis at its vertex;

$D < 0$ — the parabola does not meet the $x$-axis.

Note: if $D = 0$, then $f(x) = a(x - r)^2$, so the minimum (or maximum) value of the function equals $0$. Example: $f(x) = x^2 + 2x + 1 = (x+1)^2$ touches the $x$-axis at $(-1, 0)$.Example 1: Minimum and Maximum ValuesProblem. Find the minimum value of $f(x) = x^2 - 4x + 3$ and the maximum value of $g(x) = -x^2 + 5x - 6$.Solution.For $f(x) = x^2 - 4x + 3$: here $a = 1 > 0$, so the function has a minimum.$$r = -\frac{b}{2a} = -\frac{-4}{2 \cdot 1} = 2, \qquad f(2) = 4 - 8 + 3 = -1$$So $\min f = -1$.For $g(x) = -x^2 + 5x - 6$: here $a = -1 < 0$, so the function has a maximum.$$r = -\frac{5}{2 \cdot (-1)} = \frac{5}{2}, \qquad g\left(\frac{5}{2}\right) = -\frac{25}{4} + \frac{25}{2} - 6 = \frac{1}{4}$$So $\max g = \frac{1}{4}$.Answer: $\min f = -1$, $\max g = \frac{1}{4}$.Example 2: Parameter from a Maximum ValueProblem. The maximum value of $f(x) = -x^2 + 8x + n$ is $3$. Find $n$.Solution.Since $a = -1 < 0$, the function indeed has a maximum, attained at$$r = -\frac{b}{2a} = -\frac{8}{2 \cdot (-1)} = 4$$The maximum value is$$f(4) = -16 + 32 + n = 16 + n$$By the condition $16 + n = 3$, hence$$n = -13$$Answer: $n = -13$.The same scheme works for any problem of the type "$\max f = 4$, find $m$": find $r$, substitute, equate $f(r)$ to the given value.Example 3: Parameter from the Vertex PositionProblem. The vertex of the parabola $f(x) = mx^2 - (2m - 8)x + 1$ is at $x = -1$. Find $m$.Solution.Here $a = m$ and $b = -(2m - 8)$. The abscissa of the vertex is$$r = -\frac{b}{2a} = \frac{2m - 8}{2m}$$By the condition $r = -1$:$$\frac{2m - 8}{2m} = -1 \;\Rightarrow\; 2m - 8 = -2m \;\Rightarrow\; 4m = 8 \;\Rightarrow\; m = 2$$Answer: $m = 2$.Check: with $m = 2$ we get $f(x) = 2x^2 + 4x + 1$ and $r = -\frac{4}{4} = -1$. Correct. If the problem also asked for the ordinate of the vertex $T(-1, k)$, we would compute $k = f(-1) = 2 - 4 + 1 = -1$.Example 4: Maximum of a ProductProblem. Given $A = 2x - 12$ and $B = -2x + 24$, find the maximum value of $A \cdot B$.Solution.The product is a quadratic function of $x$:$$A \cdot B = (2x - 12)(-2x + 24) = -4x^2 + 72x - 288$$Since $a = -4 < 0$, the product has a maximum. Its roots are $x_1 = 6$ (from $A = 0$) and $x_2 = 12$ (from $B = 0$), so the vertex is at the midpoint:$$r = \frac{6 + 12}{2} = 9$$At $x = 9$: $A = 2 \cdot 9 - 12 = 6$ and $B = -2 \cdot 9 + 24 = 6$, therefore$$\max(A \cdot B) = 6 \cdot 6 = 36$$Answer: $36$.Shortcut: $A + B = 12$ is constant, so the maximum of the product occurs when $A = B = 6$, giving $36$ immediately. In combined problems like $\max(B) + \min(A)$ on a given interval, evaluate each quantity separately using the same vertex technique or the interval endpoints.Example 5: Tangency of a Line and a ParabolaProblem. The line $y = 3x + 1$ is tangent to the parabola $y = x^2 - kx + 10$. Find the sum of all possible values of $k$.Solution.Set the expressions equal:$$x^2 - kx + 10 = 3x + 1 \;\Rightarrow\; x^2 - (k + 3)x + 9 = 0$$Tangency means exactly one common point, i.e. $D = 0$:$$D = (k + 3)^2 - 4 \cdot 1 \cdot 9 = (k + 3)^2 - 36 = 0$$Hence $(k + 3)^2 = 36$, so $k + 3 = 6$ or $k + 3 = -6$, giving$$k_1 = 3, \qquad k_2 = -9$$The sum of all values: $k_1 + k_2 = 3 + (-9) = -6$.Answer: $-6$.All course topics · Practice problemsPlease enable JavaScript to use the practice app.