Topic 2

Combining Systems, Functions, and Graphs

This topic brings together three core algebra domains: systems of equations, functions and their properties, and coordinate-plane graphs. On the SAT, harder problems frequently mix these areas. You might need to solve a system of equations to find a parameter that appears in a function rule, or interpret the intersection of two graphs as the solution to a system.

Systems meet functions: A common pattern is two functions

f(x)

and

g(x)

whose graphs intersect. Setting

f(x) = g(x)

gives you an equation to solve, and the solution is the

x

-coordinate of the intersection point. If the functions are linear, you get a system of two linear equations. If one or both are quadratic, solving

f(x) = g(x)

becomes a quadratic equation, and the number of solutions tells you whether the graphs intersect zero, one, or two times.

Functions meet graphs: The SAT often describes a function with an equation and then asks about features of its graph: intercepts, vertex, slope, or end behavior. Going the other direction, a problem might give you a graph and ask you to write or identify the equation. Key connections to keep in mind:

The $y$ -intercept of $f(x)$ is $f(0)$ .
The $x$ -intercepts (zeros) are the solutions of $f(x) = 0$ .
The vertex of a quadratic $f(x) = ax^2 + bx + c$ is at $x = -\dfrac{b}{2a}$ .
A system of one linear and one quadratic equation can have 0, 1, or 2 solutions. Use the discriminant of the resulting quadratic to determine which.

Systems with parameters: Some problems include a constant (like

k

) in one of the equations and ask for the value of

k

that makes the system have exactly one solution, no solution, or infinitely many solutions. For two linear equations, parallel lines (no solution) have equal slopes but different intercepts. For a line and a parabola, exactly one intersection means the discriminant of the combined equation is zero.

Function composition and transformation: If

f(x) = 2x + 1

and

g(x) = x^2

, then

f(g(3)) = f(9) = 19

. The SAT tests this at a straightforward level. Transformations (shifts, reflections, stretches) connect a function's equation to how its graph moves:

f(x - 3)

shifts right by 3,

f(x) + 5

shifts up by 5, and

-f(x)

reflects over the

x

-axis.

Core Skills

Find the intersection point(s) of two functions by setting them equal and solving.
Use the discriminant to determine the number of intersections between a line and a parabola.
Evaluate composite functions and apply function notation in multi-step problems.
Connect features of a function's equation (slope, vertex, intercepts) to features of its graph.
Find the value of a parameter in a system that produces a specified number of solutions.

Example 1: Intersection of Two Linear Functions

Let

f(x) = 3x + 2

and

g(x) = -x + 10

. At what point do the graphs of

f

and

g

intersect? Step 1: Set

f(x) = g(x)

3x + 2 = -x + 10 \implies 4x = 8 \implies x = 2

Step 2: Find the

y

-coordinate.

f(2) = 3(2) + 2 = 8

The intersection point is

\boxed{(2, 8)}

Example 2: Line and Parabola System

For what value of

k

does the line

y = 2x + k

intersect the parabola

y = x^2

at exactly one point? Step 1: Set the equations equal.

x^2 = 2x + k \implies x^2 - 2x - k = 0

Step 2: For exactly one intersection, the discriminant equals zero.

b^2 - 4ac = (-2)^2 - 4(1)(-k) = 4 + 4k = 0 \implies k = -1

The value of

k

\boxed{-1}

Example 3: Composite Functions

f(x) = x^2 - 1

and

g(x) = 2x + 3

, what is

f(g(2))

? Step 1: Evaluate the inner function first.

g(2) = 2(2) + 3 = 7

Step 2: Evaluate the outer function at that result.

f(7) = 7^2 - 1 = 49 - 1 = 48

\boxed{48}

Example 4: System with No Solution

For what value of

a

does the system below have no solution?

y = 3x + 5 \qquad y = ax - 1

Step 1: Two lines have no solution when they are parallel: same slope, different

y

-intercepts. The first line has slope 3 and

y

-intercept 5. Step 2: For the lines to be parallel,

a = 3

. Since the

y

-intercepts (

5

and

-1

) are already different,

a = 3

gives no solution. The value of

a

\boxed{3}

Example 5: Graph Features from a Quadratic

The function

f(x) = -2(x - 3)^2 + 8

is graphed in the

xy

-plane. What is the maximum value of

f

, and at what value of

x

does it occur? Step 1: The function is in vertex form

a(x - h)^2 + k

. The vertex is

(3, 8)

. Since

a = -2 < 0

, the parabola opens downward, so the vertex is a maximum. The maximum value is

\boxed{8}

, occurring at

x = \boxed{3}

Example 6: System Solved by Substitution with a Quadratic

Solve the system:

y = x + 1

and

y = x^2 - 3x + 5

. Step 1: Set the expressions equal.

x + 1 = x^2 - 3x + 5 \implies 0 = x^2 - 4x + 4 = (x - 2)^2

Step 2: Solve.

x - 2 = 0 \implies x = 2

y = 2 + 1 = 3

The system has one solution:

\boxed{(2, 3)}

Key Takeaways

The intersection of two graphs is found by setting the functions equal. The algebraic method always matches the geometric picture.
The discriminant $b^2 - 4ac$ is your tool for counting intersections between a line and a parabola. Positive gives two intersections, zero gives one (tangent), and negative gives none.
For parallel lines (no solution), match slopes; for identical lines (infinitely many solutions), match both slope and intercept.
When evaluating composite functions, always work inside-out: evaluate the innermost function first, then use that result as the input for the outer function.