Objective

Synthesize everything learned above into challenging, multi-step SAT-style problems. Most questions combine two or more concepts from previous drills.

Questions

1

If

$x^{\frac{a}{2}} \cdot \sqrt[3]{x} = x^2$

for $x>0$

what is the value of $a$?

2

The function $g$ is defined by

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

where $a$, $h$, and $k$ are constants.

If the graph of $y=g(x)$ passes through $(2,5)$ and has its vertex at $(5,-4)$, what is the value of $a$?

3

In the system below, $k$ is a constant.

If the system has exactly one real solution, what is the value of $k$?

$y=3x^2-12x+k$

$y=6x-11$

4

If

$(2x+5)(ax-3)=6x^2+bx-15$

for all real values of $x$

what is the value of $b$?

5

What is the product of all real values of $x$ that satisfy

$\sqrt{2x+12}=x-2$?

6

If

$x^2-kx+49=0$

has exactly one real solution, and $k>0$

what is the value of $k$?

7

If

$f(x)=3x^2-18x+22$

which equivalent form explicitly reveals the minimum value of $f(x)$ as a constant?

A) $f(x)=3(x-3)^2-5$

B) $f(x)=3(x^2-6x)+22$

C) $f(x)=3(x-1)(x-5)-7$

D) $f(x)=3x(x-6)+22$

8

If

$x^2-10x+c=0$

has no real solutions,

what is the smallest possible integer value of $c$?

9

If

$\sqrt[4]{x^a}\cdot\sqrt{x^3}=x^2$

for $x>0$

what is the value of $a$?

10

The equations

$y=x^2-4x+7$

and

$y=mx+3$

intersect at exactly one point in the first quadrant.

What is the value of $m$?

Capstone Answer Key & Step-by-Step Solutions

1

Answer: $\frac{10}{3}$

Convert everything to rational exponents:

$x^{\frac{a}{2}}\cdot x^{\frac{1}{3}}=x^2$

Match exponents:

$\frac{a}{2}+\frac{1}{3}=2$

$\frac{a}{2}=\frac{5}{3}$

$a=\frac{10}{3}$

2

Answer: $1$

Use the vertex:

$y=a(x-5)^2-4$

Substitute the point $(2,5)$:

$5=a(2-5)^2-4$

$5=9a-4$

$9=9a$

$a=1$

3

Answer: $16$

Set the equations equal:

$3x^2-12x+k=6x-11$

$3x^2-18x+(k+11)=0$

Exactly one real solution means:

$b^2-4ac=0$

$(-18)^2-4(3)(k+11)=0$

$324-12(k+11)=0$

$324-12k-132=0$

$192=12k$

$k=16$

4

Answer: $9$

Match the leading coefficient:

$2a=6$

$a=3$

Substitute into the product:

$(2x+5)(3x-3)$

Middle term:

• $-6x+15x=9x$

Therefore:

$b=9$

5

Answer: $6$

Square both sides:

$2x+12=x^2-4x+4$

$x^2-6x-8=0$

Potential solutions:

$x=3\pm\sqrt{17}$

Check the solutions in the original equation.

Only

$x=3+\sqrt{17}$

is valid.

Therefore the product of all valid real solutions is:

$3+\sqrt{17}$

6

Answer: $14$

Exactly one real solution means:

$b^2-4ac=0$

$(-k)^2-4(1)(49)=0$

$k^2-196=0$

$k=14$

since $k>0$.

7

Answer: A

Vertex form:

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

directly reveals the minimum value.

Choice A shows:

$f(x)=3(x-3)^2-5$

Minimum value:

• $-5$

8

Answer: $26$

No real solutions means:

$b^2-4ac<0$

$(-10)^2-4(1)(c)<0$

$100-4c<0$

$c>25$

Smallest integer:

$26$

9

Answer: $2$

Convert to exponents:

$x^{\frac{a}{4}}\cdot x^{\frac{3}{2}}=x^2$

Match exponents:

$\frac{a}{4}+\frac{6}{4}=\frac{8}{4}$

$a+6=8$

$a=2$

10

Answer: $0$

Set the equations equal:

$x^2-4x+7=mx+3$

$x^2-(4+m)x+4=0$

Exactly one intersection means:

$b^2-4ac=0$

$[-(4+m)]^2-4(1)(4)=0$

$(4+m)^2=16$

$4+m=\pm4$

$m=0$

or

$m=-8$

Check the intersection location:

If $m=0$:

$x=2$

which lies in Quadrant I.

If $m=-8$:

$x=-2$

which lies in Quadrant II.

Therefore:

$m=0$