Algebra I: Quadratics

How to do it:

1. Write $ax^2+bx+c=0$ with $a \neq 0\text{.}$

2. Factor into $(px+q)(rx+s)$ when you can.

3. Set each factor equal to 0. Those $x$-values are the zeros of $f(x)=ax^2+bx+c\text{.}$

Worked example: $x^2-5x+6=0$ factors as $(x-2)(x-3)=0\text{,}$ so $x=2$ or $x=3\text{.}$ Check in difference of squares when the middle term vanishes.

Check: expanding must return the original. $(x-2)(x-3)=x^2-5x+6\text{.}$ The graph crosses the $x$-axis at those zeros.

How to do it:

1. Identify $a\text{,}$ $b\text{,}$ and $c$ from $ax^2+bx+c=0\text{.}$

2. Compute the discriminant $D=b^2-4ac\text{.}$

3. $x = \dfrac{-b \pm \sqrt{D}}{2a}\text{.}$ Two real roots if $D>0\text{,}$ one if $D=0\text{,}$ none real if $D<0\text{.}$

Worked example: $x^2-5x+6=0$ has $a=1,b=-5,c=6\text{.}$ $D=1\text{,}$ $x=\dfrac{5\pm 1}{2}\text{,}$ so $3$ and $2\text{.}$ Replay Quadratic formula.

Check: a negative discriminant is not "no solution" in the complex world; in Algebra I we say no real solutions. Keep the $\pm\text{:}$ dropping it loses a root.

How to do it:

1. If $a \neq 1\text{,}$ divide through by $a$ first.

2. Move the constant. Add $(\tfrac{b}{2})^2$ to both sides.

3. Write a square and take square roots. Vertex form $y=(x-h)^2+k$ reads the vertex $(h,k)\text{.}$

Worked example: $x^2-6x-7=0$ → $x^2-6x=7$ → add 9: $(x-3)^2=16$ → $x-3=\pm 4$ → $x=7$ or $x=-1\text{.}$

Check: the vertex is $(3,-16)$ for $y=(x-3)^2-16\text{,}$ not $(3,16)\text{.}$ See vertex formula.

An intercept is named for the axis it crosses, and on that axis the other variable is $0\text{.}$ So an x-intercept has $y=0$ and a y-intercept has $x=0\text{.}$ That one rule settles which variable to zero out.

How to do it:

1. y-intercept: set $x=0\text{.}$ In $y=ax^2+bx+c$ that leaves $y=c\text{,}$ so the y-intercept is the constant term $(0,c)\text{.}$ It is a substitution, so there is exactly one.

2. x-intercepts: set $y=0$ and solve $ax^2+bx+c=0$ by factoring, square roots, or the formula. There can be $0\text{,}$ $1\text{,}$ or $2\text{,}$ because a parabola can miss the axis.

3. Line of symmetry: average the two x-intercepts, $x=\dfrac{x_1+x_2}{2}\text{.}$ It is a vertical line, so write it as an equation like $x=-4\text{,}$ not as a point.

4. Vertex: put that $x$ back into the equation to get $y\text{.}$ The vertex is the turning point, a minimum if $a>0$ (opens up) and a maximum if $a<0$ (opens down).

Worked example: $y=x^2+8x+7\text{.}$

• y-intercept: set $x=0\text{,}$ giving $(0,7)\text{,}$ the constant term.

• x-intercepts: set $y=0$ and factor $(x+7)(x+1)=0\text{,}$ giving $(-7,0)$ and $(-1,0)\text{.}$

• line of symmetry: $\dfrac{-7+(-1)}{2}=-4\text{,}$ so $x=-4\text{.}$

• vertex: $y=(-4)^2+8(-4)+7=-9\text{,}$ giving $(-4,-9)\text{.}$ Since $a=1>0\text{,}$ it is a minimum at $(-4,-9)$.

Check: the line of symmetry is a line, so the answer is $x=-4\text{;}$ writing just $-4$ or the point $(-4,-9)$ answers a different question. And an x-intercept means you are crossing the x-axis, where every point has $y=0\text{;}$ the most common slip is zeroing the wrong variable.

How to do it:

1. A y-intercept is a substitution, so there is exactly one: a function has one output at $x=0\text{.}$ Replace every $x$ with $0$ and simplify. Fast.

2. x-intercepts need real solving, so there can be $0\text{,}$ $1\text{,}$ or $2$: a parabola, or a V, can miss the axis entirely. Replace $y$ with $0\text{,}$ then solve what is left by factoring, square roots, or splitting an absolute value into two cases.

3. From a graph, read them off: find where the curve touches or crosses an axis and write the ordered pair, never a bare number. Passing near a grid point is not an intercept; it must land on the axis.

4. Write the answers as points: x-intercepts $(x,0)$ with the zero second, the y-intercept $(0,y)$ with the zero first.

Worked example: the same four-step read works for a V. Take $y=|x+2|-1\text{.}$

• y-intercept: set $x=0\text{:}$ $y=|0+2|-1=1\text{,}$ giving $(0,1)\text{.}$

• x-intercepts: set $y=0$ and get the bars alone: $|x+2|=1\text{.}$ An absolute value splits into two cases, $x+2=1$ or $x+2=-1\text{,}$ so $x=-1$ or $x=-3\text{,}$ giving $(-1,0)$ and $(-3,0)\text{.}$

• line of symmetry: the fold runs through the vertex, halfway between the x-intercepts: $\dfrac{-3+(-1)}{2}=-2\text{,}$ so $x=-2\text{.}$

• vertex: $y=|-2+2|-1=-1\text{,}$ giving $(-2,-1)\text{,}$ a minimum. The arms have no endpoints; both rise forever.

Check: the $+2$ inside the bars shifts the graph left $2\text{,}$ not right; whatever is grouped with $x$ moves the graph the opposite way. And the discriminant is the same idea for a parabola: $D<0$ is the "$0$ x-intercepts" case, $D=0$ is one (the vertex sits on the axis), and $D>0$ is two.