Quadratic Functions and Equations: Chapter Test

Match the template $a(x - h)^2 + k$. The inside reads $x + 9 = x - (-9)$, so $h = -9$, while the outside gives $k = 4$ and $a = -5$.

$\text{vertex } (-9, 4), \qquad a = -5 < 0$

A negative leading coefficient opens the parabola downward, so the vertex is the highest point of the graph and $4$ is the maximum value.

The left side is a product of the two real numbers $3x + 8$ and $x - 6$, and it equals zero, so the Zero Product Property forces one of them to be zero.

$3x + 8 = 0 \ \Rightarrow \ x = -\frac{8}{3}, \qquad x - 6 = 0 \ \Rightarrow \ x = 6$

The solution set is the union of the two linear solution sets, so both values are kept.

Read $a = 4$, $b = 5$, $c = 7$ and compute the discriminant before solving anything.

$\Delta = 5^2 - 4(4)(7) = 25 - 112 = -87 < 0$

A negative discriminant puts the vertex clear of the $x$-axis on the side the parabola never returns from, so the curve never meets the axis and there is no real solution. That is a complete answer, not a stalled computation.

With $a = 4$, $b = -9$ and $c = -7$, both totals come straight from the coefficients and nothing is solved.

$r_1 + r_2 = -\frac{b}{a} = -\frac{-9}{4} = \frac{9}{4}, \qquad r_1 r_2 = \frac{c}{a} = -\frac{7}{4}$

Only the sum carries the extra minus sign; the product is $c$ over $a$, with no sign change of its own.

Half of $14$ is $7$, and $7^2 = 49$ is the one constant that completes the square. Add it and subtract it in the same line, so the value is untouched.

$x^2 + 14x + 5 = \left(x^2 + 14x + 49\right) - 49 + 5 = (x + 7)^2 - 44$

Expanding $(x + 7)^2 - 44$ returns $x^2 + 14x + 49 - 44$, which is the original expression.

The product is already compared to zero and already factored, so its roots are $x = 7$ and $x = -5$, and they cut the number line into three pieces.

$(x - 7)(x + 5) < 0 \quad \Longleftrightarrow \quad -5 < x < 7$

The leading coefficient is positive, so the parabola opens upward and is negative exactly between its roots. The symbol is strict, so the roots themselves are excluded and the interval is open.

Factored form displays the zeros: $y = 0$ at $x = 3$ and at $x = -11$. Two inputs sharing an output are mirror partners, so the axis of symmetry stands midway between them.

$x = \frac{3 + (-11)}{2} = \frac{-8}{2} = -4$

The vertex sits on the axis, so its $x$-coordinate is $-4$. The leading coefficient $2$ changes how high or low the vertex sits, never where it sits left to right.

Read $a = 2$, $b = -7$, $c = -5$, and substitute in parentheses so the squares and double negatives take care of themselves. A negative $c$ makes the $-4ac$ term add.

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(-5)}}{2(2)} = \frac{7 \pm \sqrt{49 + 40}}{4} = \frac{7 \pm \sqrt{89}}{4}$

Since $89$ is not a perfect square the roots are irrational, which is exactly why no integer pair search could have reached them.

Two crossings are two inputs sharing the output $0$, so their average is the axis of symmetry, and a standard form puts the axis at $-\tfrac{b}{2a}$.

$-\frac{b}{2(5)} = 4 \quad \Longrightarrow \quad b = -40$

The totals give the same answer: the crossings add to $2 \times 4 = 8$, and $-\tfrac{b}{a} = 8$ with $a = 5$ again gives $b = -40$. The constant $c$ is never needed, because it moves the curve up and down without moving its axis.

With a leading coefficient other than $1$, the pair to hunt for multiplies to $ac = 10 \times (-6) = -60$ and adds to $11$, which is $15$ and $-4$. Split the middle term and group.

$10x^2 + 15x - 4x - 6 = 5x(2x + 3) - 2(2x + 3) = (5x - 2)(2x + 3) = 0$

The Zero Product Property then gives $5x - 2 = 0$ or $2x + 3 = 0$, so the solution set is $\left\{-\tfrac{3}{2}, \tfrac{2}{5}\right\}$, with both elements kept.

Make the leading coefficient positive first, remembering that multiplying an inequality by a negative number reverses it.

$-x^2 + 3x + 18 \ge 0 \ \Longleftrightarrow \ x^2 - 3x - 18 \le 0 \ \Longleftrightarrow \ (x - 6)(x + 3) \le 0$

The rewritten parabola opens upward, so it is at or below zero on the closed interval between its roots $-3$ and $6$. Reading the original downward parabola directly agrees: it is at or above zero between those same two roots.

Compute all three discriminants, then ask separately of each whether it is a perfect square.

$\Delta_u = 81 - 20 = 61, \qquad \Delta_v = 81 - 72 = 9 = 3^2, \qquad \Delta_w = 81 + 20 = 101$

All three are positive, so all three have two real roots and all three factor over the reals. Only $\Delta_v$ is a perfect square, so only $v$ has rational roots, and only $v$ factors with rational coefficients: $v(x) = (2x + 3)(x + 3)$. The other two factor over the reals with irrational coefficients, which is a different claim about a different number system.

Take the $2$ out of the $x^2$ and $x$ terms only; the constant carries no $x$ and stays outside. Inside, half of $-10$ is $-5$ and $(-5)^2 = 25$.

$2x^2 - 20x + 9 = 2\left(x^2 - 10x\right) + 9 = 2\left((x - 5)^2 - 25\right) + 9 = 2(x - 5)^2 - 41$

The $-25$ sits inside the bracket, so distributing the $2$ turns it into $-50$, and $-50 + 9 = -41$.

Completing the square in general puts the vertex height at $-\tfrac{\Delta}{4a}$, so read the two signs one at a time.

$\Delta < 0 \ \text{ and } \ a < 0 \quad \Longrightarrow \quad -\frac{\Delta}{4a} = -\frac{(\text{negative})}{(\text{negative})} < 0$

So the vertex lies below the $x$-axis. The parabola opens downward, so that vertex is the highest point of the whole curve, and everything else is lower still: the graph never reaches the axis, which is what $\Delta < 0$ says on its own.

No real solution is exactly the case $\Delta < 0$, and $p$ enters the discriminant once, with the leading coefficient beside it.

$\Delta = (-4)^2 - 4(7)p = 16 - 28p < 0 \quad \Longrightarrow \quad 28p > 16 \quad \Longrightarrow \quad p > \frac{4}{7}$

At $p = \tfrac{4}{7}$ the discriminant is zero and the parabola is tangent to the axis, which is one real solution rather than none, so that single value is excluded.

The quadratic is monic, so the roots add to $11$ and multiply to $7$, both read straight off the coefficients. Expanding a square supplies the identity that rewrites the target in those two totals.

$r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1r_2 = 11^2 - 2(7) = 121 - 14 = 107$

The roots themselves are irrational, since $\Delta = 121 - 28 = 93$ is not a perfect square, and they were never needed.

Check the discriminant before hunting for roots.

$\Delta = 5^2 - 4(-3)(-9) = 25 - 108 = -83 < 0$

There is no real root, so the quadratic never changes sign, and which sign it keeps is settled by $a = -3 < 0$: the parabola opens downward with its vertex at height $-\tfrac{\Delta}{4a} = -\tfrac{83}{12} < 0$, so every output is negative. Every real number therefore satisfies the strict inequality, and the solution set is $\mathbb{R}$.

With $a = 3$, $b = -8$ and $c = -5$, the numerator opens with $-b = 8$, and a negative $c$ makes the $-4ac$ term add.

$x = \frac{8 \pm \sqrt{(-8)^2 - 4(3)(-5)}}{2(3)} = \frac{8 \pm \sqrt{124}}{6} = \frac{8 \pm 2\sqrt{31}}{6} = \frac{4 \pm \sqrt{31}}{3}$

The radical simplifies because $124 = 4 \times 31$, and the factor of $2$ then cancels from every term of the fraction, the radical term included.

When the leading coefficient is not $1$, the pair being hunted for multiplies to $ac$, not to $c$: here $ac = 6 \times 12 = 72$, and $-8$ and $-9$ do the job.

$6x^2 - 8x - 9x + 12 = 2x(3x - 4) - 3(3x - 4) = (3x - 4)(2x - 3)$

The formula reaches the same two roots with no search at all, since $\sqrt{289 - 288} = 1$ gives $x = \tfrac{17 \pm 1}{12}$, that is $\tfrac{3}{2}$ and $\tfrac{4}{3}$. A search that comes up empty reports on the list that was searched; here the list was the wrong one.

Two crossings fix the factors and leave the leading coefficient free, so write $y = a(x + 6)(x - 2)$ and let the third point decide $a$.

$-21 = a(1 + 6)(1 - 2) = -7a \quad \Longrightarrow \quad a = 3$

The parabola is $y = 3(x + 6)(x - 2)$, and its $y$-intercept is its output at $x = 0$, namely $3(6)(-2) = -36$. Without the third point every member of the family would have been an equally good answer.