Parabolas: Free Response

Both conclusions come from the leading coefficient $a$, and only from it. Here $a=-4$.

$a = -4 < 0 \;\Rightarrow\; \text{opens downward}.$

Width compares $\lvert a \rvert$ to $1$, the coefficient in $y=x^2$:

$\lvert -4 \rvert = 4 > 1 \;\Rightarrow\; \text{narrower than } y=x^2.$

Both readings use $a$ alone; $b$ and $c$ never enter either question.

Set $x=0$; the $x^2$ and $x$ terms vanish, leaving the constant term.

$y = -4(0)^2 + (0) - 2 = -2.$

So the y-intercept is $(0,-2)$.

Direction and width in part A came only from $a=-4$; the computation never used $b$. The y-intercept in part B came only from setting $x=0$, which erases the $bx$ term entirely and leaves $c=-2$; again $b$ never entered.

So changing $b$ to any other number changes neither answer. The one formula in this lesson where $b$ actually appears is the axis of symmetry,

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

and through it the vertex: a different $b$ slides the whole curve left or right along the x-axis while its shape (from $a$) and its y-intercept (from $c$) stay fixed.

Use $x=-\tfrac{b}{2a}$ with $a=2$, $b=-12$.

$x = -\frac{-12}{2(2)} = \frac{12}{4} = 3.$

So the axis of symmetry is the line $x=3$.

The y-intercept sits at $x=0$, a distance of $3$ units to the left of the axis $x=3$. Its mirror point sits the same distance, $3$ units, on the other side of the axis.

$3 + 3 = 6.$

Since a mirror pair always shares the same height, the point is $(6,7)$, with no new substitution needed.

Writing $y=ax^2+bx+c$ in completed-square form gives

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

and $x$ appears only inside that one squared term. A point $d$ units to one side of the axis has $x+\tfrac{b}{2a}=d$; a point $d$ units to the other side has $x+\tfrac{b}{2a}=-d$. Squaring erases the sign, so both give $\left(\pm d\right)^2=d^2$ inside the formula, and therefore the identical value of $y$. That is exactly why reflecting across the axis, as in part B, always finds a genuine second point rather than a guess.

Vertex form is $y=a(x-h)^2+k$ with vertex $(h,k)$. Matching term by term, $x-5$ gives $h=5$ and the loose constant gives $k=-2$.

$(h,k) = (5,-2).$

Rewrite the plus sign as vertex form expects it: $x+7=x-(-7)$, so $h=-7$, and $k=4$.

$(h,k) = (-7,4).$

The number written inside the parentheses, $7$, is the OPPOSITE of the vertex's x-coordinate.

The template is fixed as $y=a(x-h)^2+k$, with a MINUS sign built into it. Whatever is actually written inside the parentheses is $x-h$; if it appears on the page as $x+p$ for some positive number $p$, the only way to match the template is

$x+p = x-(-p),$

which forces $h=-p$. Since $p>0$, $h=-p$ is negative. So a plus sign inside the parentheses is never a coincidence, it is the algebra insisting that $h$ be negative, and the vertex's x-coordinate is exactly $h$.

One counterexample is enough to break a claim about EVERY parabola, so pick any quadratic and check it honestly.

Take $y=2x^2-4x+7$, so $a=2$, $b=-4$, $c=7$.

$D = b^2-4ac = (-4)^2 - 4(2)(7) = 16 - 56 = -40.$

Since $D<0$ there is no real solution to $2x^2-4x+7=0$, so this parabola has zero x-intercepts, not two. The claim is false as stated.

Compute the discriminant with $a=1$, $b=-10$, $c=25$.

$D = (-10)^2 - 4(1)(25) = 100 - 100 = 0.$

A zero discriminant means exactly one real x-intercept, and geometrically that happens when the vertex itself sits on the x-axis rather than above or below it. Here the vertex is at $x=-\tfrac{-10}{2}=5$, and substituting back gives $y=(5)^2-10(5)+25=0$, confirming the vertex $(5,0)$ is that one intercept.

The three cases come from the quadratic formula,

$x = \frac{-b \pm \sqrt{D}}{2a},$

and what its $\pm\sqrt{D}$ can do: $D>0$ gives two distinct real roots, so two x-intercepts; $D=0$ collapses the $\pm$ to one value, so one x-intercept, the vertex sitting on the axis; $D<0$ gives no real root at all, so no x-intercepts.

Restricting to $a>0$ (upward-opening) does not rescue the 'always two' claim. Part A's own counterexample, $y=2x^2-4x+7$, already has $a=2>0$ and still has zero x-intercepts, because its upward-opening vertex happens to sit above the x-axis. Nothing about opening upward forces the vertex below the axis; the sign of $D$ is decided by $a$, $b$, AND $c$ together, not by the sign of $a$ alone. So the restricted version fails for exactly the same shape of counterexample already in hand.

Substitute $x=-\tfrac{b}{2a}$ into $y=ax^2+bx+c$ and simplify each term in turn.

$y = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c.$

The first term:

$a\cdot\frac{b^2}{4a^2} = \frac{b^2}{4a}.$

The second term:

$b\cdot\left(-\frac{b}{2a}\right) = -\frac{b^2}{2a}.$

Add the two together with the third:

$y = \frac{b^2}{4a} - \frac{b^2}{2a} + c = \frac{b^2}{4a} - \frac{2b^2}{4a} + c = -\frac{b^2}{4a} + c = c - \frac{b^2}{4a}.$

This is exactly the vertex height, reached this time by direct substitution rather than by completing the square.

Read off $a=3$, $b=-6$, $c=8$ and substitute directly into $c-\tfrac{b^2}{4a}$.

$c - \frac{b^2}{4a} = 8 - \frac{(-6)^2}{4(3)} = 8 - \frac{36}{12} = 8 - 3 = 5.$

The vertex height is $5$, found without ever computing the vertex's x-coordinate.

The y-intercept is $c$; the vertex height, from part A, is $c-\tfrac{b^2}{4a}$. Subtract the second from the first:

$c - \left(c-\frac{b^2}{4a}\right) = \frac{b^2}{4a}.$

Since $b^2\ge 0$ for every real $b$, and $a>0$ is given, this difference is never negative:

$\frac{b^2}{4a} \ge 0.$

So the y-intercept can never sit below the vertex height, only above it or equal to it, for every upward-opening parabola, not just the ones tried by example. Equality happens exactly when the difference is $0$, which requires $b^2=0$, i.e. $b=0$: the case where the axis of symmetry is the y-axis itself and the y-intercept IS the vertex.