If you want to get better at solving quadratic equations on the SAT, understanding functions is where it all connects. Functions appear across every section of SAT Math, from linear models to quadratic formula problems to exponential growth and decay. Knowing how to solve a quadratic equation using multiple methods, recognising when completing the square to solve is the fastest path, and applying the discriminant formula to answer questions without fully solving them are skills that show up in question after question.
This guide walks through every function type of the SAT tests, with challenging worked examples and clear strategies for each concept.
What are Linear, Quadratic and Exponential Functions?
Watch this video to learn how to identify linear, quadratic, and exponential functions by analyzing how the output changes with each step. This is a key skill for SAT Math functions questions.
How to Solve Quadratic Equations on the SAT
Three methods are commonly tested. Factoring is fastest when coefficients and roots are integers. The quadratic formula works on any quadratic in the form ax² + bx + c = 0 and never fails. Completing the square is most useful for vertex form conversions and minimum or maximum value problems. Before choosing a method, calculate the discriminant formula b² – 4ac. A positive discriminant means two real solutions, zero means one repeated solution, and a negative discriminant means no real solutions at all.
Function Notation Reading, Evaluating, and Composing Functions
Function notation uses f(x) to name a rule that assigns exactly one output to each input. Read f(x) as “f of x”. The letter f names the function, and x is the input going in.
f(x) represents the output when x is used as the input
To evaluate a function, replace every occurrence of x with the given input and simplify completely.
Example 1. Evaluating with an algebraic expression as input
Given f(x) = 2x² – 5x + 3, find f(x + h) – f(x) and simplify fully. Then find the value of the expression when h = 0.
f(x + h) = 2(x + h)² – 5(x + h) + 3 = 2x² + 4xh + 2h² – 5x – 5h + 3
Subtract f(x)
[2x² + 4xh + 2h² – 5x – 5h + 3] – [2x² – 5x + 3]
Simplify
4xh + 2h² – 5h = h(4x + 2h – 5)
f(x + h) – f(x) = h(4x + 2h – 5)
When h = 0, the expression equals 0
Common trap – Do not cancel h before fully expanding (x + h)². Expand first, then subtract term by term.
Example 2. Composition with an unknown constant
Functions f and g are defined by f(x) = 3x – k and g(x) = x² + 2. If f(g(2)) = 17, find k. Then find all x such that g(f(x)) = 38.
g(2) = 2² + 2 = 6
f(g(2))
f(6) = 3(6) – k = 18 – k = 17
k = 1
So f(x) = 3x – 1
g(f(x)) = 38
Substitute f(x)
(3x – 1)² + 2 = 38
(3x – 1)² = 36
3x – 1 = 6 or 3x – 1 = -6
x = 7/3 or x = -5/3
Key tip – For composition, always evaluate the inner function first and carry the result into the outer function. When the composition equals a constant, set up the equation at that final step and solve using the quadratic formula or factoring.
Linear Functions Slope as Rate of Change in Context
A linear function produces a straight line. On the SAT, linear function questions almost always give a real-world context where the slope is a rate per unit and the y-intercept is a starting amount.
f(x) = mx + b
m is the slope, the amount the output changes for each one-unit increase in x. b is the y-intercept, the output when x = 0.
Example 1. System of linear functions in context
A factory produces two products. Product A costs $14 per unit to make with a fixed daily overhead of $320. Product B costs $9 per unit with a fixed daily overhead of $545. On a given day the factory spends the same total amount on both products. How many units of each product are made that day, given that the total number of units produced is 95?
Let a = units of A, b = units of B
Equation 1 a + b = 95
Equation 2 14a + 320 = 9b + 545 → 14a – 9b = 225
From eq. 1 a = 95 – b → 14(95 – b) – 9b = 225
Expand 1330 – 14b – 9b = 225 → -23b = -1105 → b = 48
a = 95 – 48 = 47
Product A – 47 units, Product B – 48 units
Example 2. Finding a slope when the model has an additional variable
The function T(d) = md + 8 models the temperature in degrees Celsius at a depth of d meters below the ocean surface. At d = 50, T = 19. At d = 200, T = 4. Find m, interpret it in context, and find the depth at which the temperature is exactly 0.
Step 1 – m = (4 – 19) / (200 – 50) = -15 / 150 = -0.1
Equation T(d) = -0.1d + 8
m = -0.1 means the temperature decreases by 0.1°C for every 1 metre increase in depth.
Set T = 0 -0.1d + 8 = 0 → d = 80
Temperature reaches 0°C at a depth of 80 metres.
Key tip – When a SAT linear model question gives two data points, always calculate slope first using the slope formula before writing the equation. Do not guess the y-intercept from the context.
Quadratic Functions Vertex Form, Standard Form, and Reading Graphs
Quadratic functions produce parabolas and appear in both algebra and graph-reading questions on the SAT. The method of completing the square to solve connects directly to vertex form, and the discriminant formula tells you about the graph’s roots without solving anything.
Standard form f(x) = ax² + bx + c
Vertex form – f(x) = a(x – h)² + k
Vertex at (h, k)
Example 1. Finding a constant then applying the discriminant
The function f(x) = 2x² – 12x + k has a minimum value of 5. Find k. Then use the discriminant formula to determine how many real solutions f(x) = 3 has.
Axis of symmetry x = -b / 2a = 12 / 4 = 3
Minimum value f(3) = 2(9) – 36 + k = -18 + k → -18 + k = 5 → k = 23
f(x) = 2x² – 12x + 23
Set f(x) = 3 2x² – 12x + 23 = 3 → 2x² – 12x + 20 = 0 → x² – 6x + 10 = 0
Discriminant (-6)² – 4(1)(10) = 36 – 40 = -4 < 0
No real solutions. The horizontal line y = 3 lies below the parabola’s minimum of 5.
Example 2. Converting to vertex form and finding all features
Convert g(x) = 3x² – 24x + 21 to vertex form using completing the square to solve for the vertex. Then find the x-intercepts using the quadratic formula and state the range.
Factor out 3 g(x) = 3(x² – 8x) + 21
Complete the square g(x) = 3(x² – 8x + 16 – 16) + 21 = 3(x – 4)² – 48 + 21
g(x) = 3(x – 4)² – 27 Vertex = (4, -27)
Quadratic formula x = (24 ± √(576 − 252)) / 6 = [24 +/- sqrt(324)] / 6 = [24 +/- 18] / 6
x = 7 or x = 1 x-intercepts at (1, 0) and (7, 0)
Range [-27, ∞) since a = 3 > 0 and minimum value is -27
Common trap – When completing the square inside parentheses with a leading coefficient, remember that adding 16 inside the bracket is really adding 3 times 16 = 48 to the expression. Always subtract the same amount outside the bracket to keep the equation balanced.
Exponential Growth and Decay Identifying from Tables and Equations
Exponential functions change by multiplying by a constant ratio rather than adding a constant difference.
f(x) = a * b^x a = initial value, b = constant ratio
If b > 1 the function grows. If 0 < b < 1 the function decays. The initial value a is the output when x = 0.
Example 1. Finding parameters from two non-adjacent points
An exponential function f(x) = a * b^x passes through (2, 18) and (5, 486). Find a and b, then find the value of x when f(x) = 4374.
Divide f(5) / f(2) = b^3 = 486 / 18 = 27 → b = 3
Find a 18 = a * 3^2 = 9a → a = 2
f(x) = 2 * 3^x
Solve f(x) = 4374 2 * 3^x = 4374 → 3^x = 2187 = 3^7 → x = 7
f(x) = 4374 when x = 7
Domain and Range
Domain is the set of all valid input values. Range is the set of all possible output values. For polynomial and exponential functions with positive bases, the domain is all real numbers. Restrictions appear with square roots and denominators.
Example 1. Domain with two simultaneous restrictions
Find the domain of h(x) = sqrt(2x – 6) / (x² – x – 12).
Restriction 1 (sqrt) 2x – 6 ≥ 0 → x ≥ 3
Restriction 2 (denom) x² – x – 12 = 0 → (x-4)(x+3) = 0 → x = 4 or x = -3
x = -3 is already excluded since x >= 3 from restriction 1
Exclude x = 4 from denominator restriction
Domain [3, 4) ∪ (4, ∞), meaning x = 3 is included, x = 4 is excluded, and all values greater than 4 are included.
Transformations Shifts, Reflections, and Stretches
Every transformation of a function follows the same structure regardless of whether the base function is linear, quadratic, or exponential.
y = a f(x – h) + k
If f(x) becomes f(bx), it causes a horizontal stretch or compression by a factor of 1/b.
h shifts horizontally (right when h > 0). k shifts vertically (up when k > 0). |a| > 1 stretches vertically. a < 0 reflects across the x-axis.
Common trap – Horizontal shifts are counterintuitive. The expression (x – h) moves the graph to the RIGHT when h is positive. This trips up students who read the minus sign and expect a left shift.
Example 1. Identifying transformation from function values alone
The table shows values of f. Function g is defined by g(x) = 2f(x + 3) – 4. Given f(1) = 5, f(4) = -2, f(7) = 3, find g(-2), g(1), and g(4).
g(-2) 2f(-2 + 3) – 4 = 2f(1) – 4 = 2(5) – 4 = 6
g(1) 2f(1 + 3) – 4 = 2f(4) – 4 = 2(-2) – 4 = -8
g(4) 2f(4 + 3) – 4 = 2f(7) – 4 = 2(3) – 4 = 2
g(-2) = 6, g(1) = -8, g(4) = 2
Example 2. Finding the equation of a transformed exponential
The graph of f(x) = 3^x is transformed so that it is reflected across the x-axis, compressed vertically by a factor of 1/2, shifted left by 2 units, and shifted up by 5. Write the equation of the transformed function g. Find the y-intercept and the horizontal asymptote of g.
Reflection across the x-axis and vertical compression by 1/2 a = -1/2
Shift left 2 replace x with (x + 2)
Shift up 5 add 5
g(x) = (-1/2) * 3^(x + 2) + 5
y-intercept g(0) = (-1/2) * 3^2 + 5 = -9/2 + 5 = 1/2
Horizontal asymptote As x → -∞, 3^(x+2) → 0, so g → 5
y-intercept = (0, 1/2), horizontal asymptote y = 5
SAT-Style Function Problems with Video Solutions
Question 1
The function f is defined by f(x) = ax² + bx + c, where the vertex is (2, -6) and the graph passes through (4, 2). A second function g is defined by g(x) = f(2x – 1). What is the y-coordinate of the y-intercept of the graph of g?
A. 4 B. 6 C. 12 D. 18
Solution
Step 1. Write f(x) in vertex form using the vertex (2, -6).
f(x) = a(x – 2)² – 6
Step 2. Find a by substituting the point (4, 2).
2 = a(4 – 2)² – 6 → 2 = 4a – 6 → 4a = 8 → a = 2
f(x) = 2(x – 2)² – 6
Step 3. Find the y-intercept of g by evaluating g(0).
g(0) = f(2(0) – 1) = f(-1)
Step 4. Evaluate f(-1).
f(-1) = 2(-1 – 2)² – 6 = 2(-3)² – 6 = 2(9) – 6 = 18 – 6 = 12
Answer C
Key Rule
For composite transformation questions, first find the full equation of the base function using vertex and a given point. Then substitute x = 0 into g(x) and work outward step by step. Never skip the intermediate f-value.
Why the other choices are wrong
A (4) results from an incorrect substitution or miscalculation in evaluating f(-1). B (6) comes from forgetting to apply the horizontal shift inside g, evaluating f(0) instead of f(-1). D (18) is the value of 2(-3)^2 before subtracting the -6 from the vertex, stopping one step too early.
Think10x.ai video explaining vertex form and function transformations (Finding the y-intercept)
Question 2
The function f is defined by f(x) = x² – 4x + 7 and g is a linear function defined by g(x) = mx + 3. The graphs of f and g are tangent to each other, meaning they intersect at exactly one point. Which of the following is a possible value of m?
A. 0 B. -4 C. 2 D. 6
Solution
Step 1. Set f(x) equal to g(x) and rearrange.
x² – 4x + 7 = mx + 3 → x² – (4 + m)x + 4 = 0
Step 2. For tangency, the graphs touch at exactly one point. Use the discriminant formula.
b² – 4ac = (4 + m)² – 4(1)(4) = 0
(4 + m)² = 16
4 + m = 4 or 4 + m = -4
m = 0 or m = -8
Step 3. Check which value appears among the choices.
m = 0 is answer choice A.
Verify m = 0 x² – 4x + 4 = 0 → (x-2)² = 0 → x = 2 (one solution, tangent confirmed)
Key rule
When two curves are tangent, they intersect at exactly one point, which means the discriminant of the combined equation equals zero. Set up the equation first by setting f(x) = g(x), rearrange to standard form, then apply b^2 – 4ac = 0.
Why the other choices are wrong
B (-4) gives discriminant = (4-4)² – 16 = -16 < 0, meaning no intersection at all. C (2) gives discriminant = 36 – 16 = 20 > 0, meaning two intersection points, not one. D (6) gives discriminant = 100 – 16 = 84 > 0, also two intersection points.
Think10x.ai video explaining tangent lines and the quadratic discriminant
Frequently Asked Questions
Work strictly from the inside out. Evaluate the inner function at the given input first, then use that result as the input to the outer function. Never try to simplify f(g(x)) algebraically when a specific number is given. Plug in immediately and calculate numerically.
Check the first differences between consecutive outputs. If those differences are constant, the relationship is linear. If they are not constant, check the ratios between consecutive outputs. If the ratios are constant, the relationship is exponential. If neither differences nor ratios are constant, the relationship is neither.
Use completing the square when the question asks for the vertex, the minimum or maximum value, or the equation in vertex form. The quadratic formula is faster when you only need the roots. Both methods always give the same roots, but completing the square gives you vertex information as a bonus, which the quadratic formula alone does not.
The discriminant b^2 – 4ac tells you how many times the parabola crosses the x-axis. A positive discriminant means two x-intercepts. A discriminant of zero means the vertex sits exactly on the x-axis, so there is one repeated root. A negative discriminant means the parabola does not cross the x-axis at all. SAT questions about the number of solutions or the number of intersections with the x-axis are always answered using the discriminant.
Look at what is happening inside the function to the x before any operations outside. In f(x – h), replacing x with x – h shifts the graph to the RIGHT by h units, even though you see a subtraction. In f(x + h), the graph shifts LEFT. The counterintuitive direction is the most common trap. Test one point to confirm direction whenever you are unsure.
Domain is all allowed input values. Range is all possible output values. For domain, look for square root restrictions (expression inside must be at least zero) and denominator restrictions (denominator cannot be zero). For range, identify the vertex of a quadratic (the minimum or maximum output), or trace what values the function can actually reach based on its shape. For exponential functions, the range is always positive when a is positive and the asymptote provides the boundary.
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