Graphing Quadratics and Inequalities: Chapter Review

Vocabulary and notation

Parabola

The smooth U-shaped curve every quadratic graphs as. It bends everywhere, so its points join into one curve, never straight segments.

Vertex

The parabola's turning point, where it stops falling and starts rising or the reverse: the one point with no mirror twin.

Axis of symmetry

The vertical line the parabola folds onto itself along: inputs equally far either side of it give the same height.

Root of a quadratic

A value of $x$ making the quadratic $0$. On the graph it is an $x$-intercept; on the number line, a boundary where the sign can change.

Circle

The set of all points a fixed distance $r$ (the radius) from a fixed point $(h, k)$ (the center). Nothing else qualifies.

Quadratic inequality

A quadratic compared to $0$ by $<$, $>$, $\le$, or $\ge$, solved for one variable, so the answer is a set on the number line, not a region.

Sign chart

The intervals the roots cut the number line into, each labeled with the quadratic's sign, from a test point or the parabola's shape.

Optimizing input and optimal value

The vertex's two coordinates, kept apart: the input is WHERE the best happens, the value is HOW GOOD it is. Rarely the same number.

Formulas and theorems

• Standard form of a quadratic

  $$y = ax^2 + bx + c$$

  Use when $a \neq 0$; with $a = 0$ the graph is the line $y = bx + c$. Read $a$, $b$, $c$ with their signs only from this form.

• What the sign and size of $a$ tell you

  $a > 0$: opens upward, vertex is a minimum. $a < 0$: opens downward, vertex is a maximum. Large $\lvert a \rvert$: narrow. Small $\lvert a \rvert$: wide.

  Use when Only the leading coefficient decides these; $b$ and $c$ do not. Width compares against $y = x^2$: $\lvert a \rvert > 1$ narrower, $\lvert a \rvert < 1$ wider.

• Axis of symmetry and the vertex

  $$x = -\frac{b}{2a}$$

  Use when $a \neq 0$. This is both the axis of symmetry and the vertex's $x$-coordinate; the $y$-coordinate comes only from substituting it back.

• Vertex form

  $$y = a(x - h)^2 + k$$

  Use when Vertex $(h, k)$, same $a$ as standard form. The form SUBTRACTS $h$, so $y = (x + 3)^2 + 2$ has $h = -3$. Matching standard form gives $h = -\tfrac{b}{2a}$, $k = c - \tfrac{b^2}{4a}$.

• The intercepts, and the discriminant that counts them

  The $y$-intercept is $(0, c)$, from setting $x = 0$; the $x$-intercepts come from setting $y = 0$ and solving $ax^2 + bx + c = 0$. $D = b^2 - 4ac$ counts them: $D > 0$ two, $D = 0$ one (the vertex on the axis), $D < 0$ none.

  Use when Zero the OTHER variable for each: always exactly one $y$-intercept, but two, one, or no $x$-intercepts. Coefficients from standard form with their signs; only real solutions land on the graph, and the count holds either way the parabola opens.

  e.g. $y = x^2 - 4x + 7$: $D = 16 - 28 = -12$, so it never meets the $x$-axis.

• Completing the square

  $$x^2 + bx + \left(\tfrac{b}{2}\right)^2 = \left(x + \tfrac{b}{2}\right)^2$$

  Use when Needs leading coefficient $1$: factor $a$ out of a quadratic's $x$-terms, or divide a circle by the shared coefficient of $x^2$ and $y^2$. In an expression add and subtract it; in an equation add it to both sides.

• Distance between two points

  $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

  Use when Any two points: the Pythagorean theorem on the two gaps, squared and added, never just added. Either point may come first, since squaring erases each difference's sign. The root is part of it, so stopping early leaves $d^2$; simplify it as a radical.

  e.g. $(-1, 2)$ to $(3, 5)$: $d = \sqrt{4^2 + 3^2} = 5$, not $4 + 3$.

• Standard (center-radius) form of a circle

  $$(x - h)^2 + (y - k)^2 = r^2$$

  Use when Center $(h, k)$, radius $r > 0$; the form SUBTRACTS each center coordinate, and the right side is $r^2$, not $r$. A right side that is not a perfect square leaves the radius as a radical. At the origin it becomes $x^2 + y^2 = r^2$, meeting the axes at $(\pm r, 0)$ and $(0, \pm r)$.

• General form of a circle

  $$x^2 + y^2 + Dx + Ey + F = 0$$

  Use when $D = -2h$, $E = -2k$, $F = h^2 + k^2 - r^2$, so the center and radius stay buried until you complete the square on each variable. The $x^2$ and $y^2$ coefficients must be EQUAL, or it is not a circle. This $D$ is the $x$-coefficient, not the discriminant.

• When a circle collapses

  For $(x - h)^2 + (y - k)^2 = R$: $R > 0$ is a circle of radius $\sqrt{R}$, $R = 0$ is the single point $(h, k)$, and $R < 0$ has no graph at all.

  Use when Completing the square on a general form always produces this shape, and $R$ is not guaranteed positive: check its sign before announcing a radius.

• Sign of a quadratic across its roots

  With $a > 0$ and distinct real roots $r < s$: positive for $x < r$, negative for $r < x < s$, positive for $x > s$. With $a < 0$ every one of those signs reverses.

  Use when Needs $D > 0$ (two distinct real roots) and one side already $0$. At a root the value is exactly $0$, which joins the solution only for $\le$ or $\ge$.

• Optimization models: area, revenue, projectile

  Fence $F$, four sides: $A = x\left(\tfrac{F}{2} - x\right)$. Fence $F$, three sides against a wall: $A = x(F - 2x)$. Revenue: $R = (\text{price})(\text{quantity})$. Height in feet: $h = -16t^2 + v_0 t + h_0$.

  Use when Each has $a < 0$, so each peaks at its vertex, the four-sided pen at a square. A length needs $0 < x < \tfrac{F}{2}$; revenue is quadratic only when the quantity falls linearly as the price rises. For height, $t$ is in seconds, $v_0$ the initial upward speed, $h_0$ the launch height, and $-4.9t^2$ replaces $-16t^2$ in metres; the vertex input is the peak's time.

  e.g. $200$ tickets at $10$ dollars, losing $10$ sales per $1$ dollar rise: $R = (10 + x)(200 - 10x)$.

Problem types, step by step

Exam traps

• Trap Reading $(x + 3)^2 + (y - 4)^2 = 36$ as center $(3, 4)$ with radius $36$.

  Fix The form subtracts each center coordinate: $x + 3$ is $x - (-3)$, so $h = -3$; the right side is $r^2$. Center $(-3, 4)$, radius $6$.

• Trap Answering $-2 < x < 3$ for $x^2 - x - 6 > 0$.

  Fix That is where the upward parabola sits BELOW the axis. Above $0$ is outside the roots: $x < -2$ or $x > 3$. Test $x = 0$, which gives $-6$.

• Trap Answering $x \le -1$ or $x \ge 3$ for $-x^2 + 2x + 3 \ge 0$, because $\ge 0$ means "outside the roots".

  Fix Outside the roots is the $a > 0$ pattern, and here $a = -1$. A downward parabola is above the axis BETWEEN its roots, so the answer is $-1 \le x \le 3$. Test $x = 0$: $3 \ge 0$ holds. Or multiply by $-1$ and reverse the symbol: $x^2 - 2x - 3 \le 0$.

• Trap Answering $3$ to "what is the minimum value of $y = x^2 - 6x + 1$".

  Fix $3$ is only where the minimum happens. Substitute back: $y = 9 - 18 + 1 = -8$, the minimum value.

• Trap Writing an outside-the-roots answer as one band, such as $3 < x < -2$.

  Fix Two separate rays joined by "or": $x < -2$ or $x > 3$, that is $(-\infty, -2) \cup (3, \infty)$. A band from the larger number down to the smaller names nothing.

• Trap Dropping the leading minus in $-\tfrac{b}{2a}$, so $y = x^2 - 4x + 1$ looks like it has its axis at $x = -2$.

  Fix The formula flips the sign of $b$: $-\tfrac{-4}{2(1)} = 2$, so the axis is $x = 2$ and the vertex $(2, -3)$.

• Trap Solving $x^2 < 4x$ by dividing both sides by $x$ to get $x < 4$.

  Fix Dividing by a variable assumes its sign and corrupts the solution set: $x = -1$ passes $x < 4$ but fails $1 < -4$. Move everything over: $x^2 - 4x < 0$, roots $0$ and $4$, so $0 < x < 4$.

• Trap Using the full perimeter when a wall supplies the fourth side, so $60$ metres of fence gives a $15$ by $15$ pen.

  Fix Three sides only: $2x + y = 60$, so $A = x(60 - 2x)$, peaking at $x = 15$ with $y = 30$, a $450$ square metre garden. The four-sided model double-counts the free wall and lands on $225$.