Quadratic equations and the parabola

A quadratic is any expression of the form $ax^{2}+bx+c$, and its graph is a parabola. The test asks you to do two kinds of work with it: solve the equation $ax^{2}+bx+c=0$ for its roots, and read features off the curve, the vertex, the intercepts, the axis of symmetry, and the highest or lowest point. Both come from the same three coefficients. The work is mechanical once you know the tools, so the points leak in four predictable places: a vertex form read with the wrong sign, a discriminant turned into the wrong count of solutions, a slip in the quadratic formula or in completing the square, and the right number reported as the wrong feature of the parabola.

Three methods solve any quadratic. Factoring is fastest when the roots are whole numbers. The quadratic formula always works and is the safe choice when factoring is not obvious. Completing the square rewrites the expression in vertex form and is what the formula is built from. Whichever you use, the two roots come from the $\pm$, so a quadratic with a positive discriminant has two solutions, not one.

• The quadratic formula: $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. The numerator starts with $-b$, and the whole thing is divided by $2a$.
• The $\pm$ produces two roots. Keeping only the plus branch drops a solution.
• Completing the square: halve $b$, square it, add and subtract that amount. The added constant is $(b/2)^{2}$, not $b/2$.

Worked example. Solve $x^{2}-6x+7=0$ with the quadratic formula.

The piece under the square root, $b^{2}-4ac$, is the discriminant, and its sign alone tells you how many real solutions the equation has. A positive discriminant means the square root is a real number you add and subtract, so there are two real solutions. A zero discriminant collapses the $\pm$ onto a single value, one repeated solution. A negative discriminant has no real square root, so there are no real solutions. The trap is inverting this: reading a negative discriminant as if it meant two solutions, or a zero as if it meant none.

• $b^{2}-4ac>0$: two distinct real solutions.
• $b^{2}-4ac=0$: exactly one real solution, a repeated root.
• $b^{2}-4ac<0$: no real solutions.

Worked example. How many real solutions does $x^{2}+2x+5=0$ have?

Vertex form is $y=a(x-h)^{2}+k$, and it hands you the vertex directly as $(h,k)$. Two things trip people. First, the sign inside the parentheses flips: $x-h$ means the vertex is at $x=h$, so $(x-3)^{2}$ sits at $x=3$ and $(x+3)^{2}$ sits at $x=-3$. Second, the coordinates have an order, $h$ first and $k$ second, and the outside constant $k$ keeps its own sign. The sign of $a$ sets the opening: positive opens up to a lowest point, negative opens down to a highest point, and either way the extreme value of $y$ is $k$.

• $y=a(x-h)^{2}+k$ has vertex $(h,k)$ and axis of symmetry $x=h$.
• The sign inside flips: $x-h$ gives $h$, so $(x+1)^{2}$ has $h=-1$.
• If $a>0$ the parabola opens up and $k$ is the minimum value of $y$; if $a<0$ it opens down and $k$ is the maximum.

Worked example. Find the vertex of $y=(x-3)^{2}+2$.

A parabola has several features, and they are different numbers from the same graph. The x-intercepts, or roots, are where $y=0$; the y-intercept is where $x=0$, which is just the constant $c$. The vertex is the turning point $(h,k)$; its x-coordinate is the axis of symmetry, halfway between the roots. The maximum or minimum value is the $y$-coordinate of the vertex, $k$, not its location. Each is a real number from the same problem, and they look alike enough to swap by accident, so the last step is always to check which one the item asked for.

Worked example. Where does $y=x^{2}-2x-8$ cross the x-axis?

Flip the sign in vertex form to read $(h,k)$, let the sign of the discriminant set the count of solutions, carry the $\pm$, the $-b$, and the $2a$ through the formula, then report the root, vertex, intercept, or value the item actually names.

Four patterns recur on quadratics and the parabola. They are the same ones the diagnostic routes on.