Quadratic Functions and Modeling
Unit 5 of Algebra I. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Graphs of quadratic functions, vertex and intercepts, comparing linear, quadratic and exponential models, transformations, square-root and cube-root functions, piecewise functions.
How this unit is tested
Start by sorting any quadratic into one of three forms — standard (ax^2+bx+c), vertex (a(x-h)^2+k), or factored (a(x-r1)(x-r2)) — because each form hands you a different piece of information for free: standard gives you the y-intercept and lets you use x=-b/(2a) for the vertex, vertex form gives you the vertex directly, and factored form gives you the x-intercepts directly. Most exam problems are really asking you to convert between these forms or to read one specific feature off whichever form you're given.
For transformations, think of every new function as g(x)=a*f(x-h)+k applied to a parent function you already know: x^2, √x, or ∛x. The value of h shifts left/right (opposite of its sign inside the parentheses), k shifts up/down, negative a reflects over the x-axis, and |a|>1 stretches vertically while 0<|a|<1 compresses. Practice describing transformations in words before you try to graph them, since the words are what exam questions usually ask for.
When comparing models from a table, don't guess from the shape of the numbers — test them. Compute first differences: if they're constant, the data is linear. If first differences aren't constant but second differences are, it's quadratic. If consecutive ratios (not differences) are constant, it's exponential. For large-x behavior, remember that exponential functions eventually overtake every polynomial function, no matter how big the polynomial's coefficients are.
For square-root and cube-root functions, always check the domain first for square roots (radicand must be ≥0) since cube roots have no such restriction and are defined for all reals. For piecewise functions, identify which interval your input falls into before you evaluate anything, paying close attention to whether the boundary is included with ≤/≥ or excluded with .
What you have to know
Vertex formula (standard form)
For f(x)=ax^2+bx+c, the vertex has x-coordinate x=-b/(2a); substitute back into f(x) to get the y-coordinate.
Vertex form
f(x)=a(x-h)^2+k has vertex (h,k), axis of symmetry x=h, and opens upward if a>0, downward if a<0.
Quadratic formula and discriminant
The solutions to ax^2+bx+c=0 are x=(-b±√(b^2-4ac))/(2a). If b^2-4ac>0 there are two real roots, if =0 one repeated real root, if <0 no real roots.
Transformation rule
For g(x)=a*f(x-h)+k, h shifts the graph of f horizontally (right if h>0, left if h<0), k shifts vertically, negative a reflects over the x-axis, and |a| scales the graph vertically.
Difference/ratio test for model type
A table represents a linear function if first differences of y are constant, a quadratic function if second differences are constant, and an exponential function if consecutive ratios of y are constant.
Piecewise function definition
A piecewise function is defined by different expressions on different, non-overlapping intervals of its domain; evaluate by first locating which interval the input belongs to, then applying only that piece's rule.
14 practice questions
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Find the vertex of f(x) = 2x^2 - 8x + 5.
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Answer. (2, -3)
Use x=-b/(2a) = 8/4 = 2. Then f(2) = 2(4) - 16 + 5 = 8 - 16 + 5 = -3, so the vertex is (2, -3). -
Rewrite g(x) = -3(x+1)^2 - 2 in standard form.
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Answer. g(x) = -3x^2 - 6x - 5
Expand (x+1)^2 = x^2+2x+1, multiply by -3 to get -3x^2-6x-3, then subtract 2 more: -3x^2-6x-5. -
A parabola has x-intercepts at x=-3 and x=1 and passes through (0,-6). Write its equation in standard form.
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Answer. f(x) = 2x^2 + 4x - 6
Start with factored form f(x)=a(x+3)(x-1). Plug in (0,-6): a(3)(-1) = -3a = -6, so a=2. Then f(x)=2(x+3)(x-1)=2x^2+4x-6. -
Which function eventually grows fastest for large x: f(x)=2^x, g(x)=100x^2, or h(x)=50x+10?
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Answer. f(x)=2^x
Exponential functions eventually exceed any polynomial function's growth, regardless of coefficients; for large enough x, 2^x will surpass both 100x^2 and 50x+10. -
A table shows x=0,1,2,3 with y=3,5,9,15. Is this data best modeled as linear, quadratic, or exponential?
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Answer. Quadratic
First differences are 2, 4, 6 (not constant). Second differences are 2, 2 (constant), which is the signature of a quadratic model. -
If g(x) = f(x-3) + 2 and f(x)=x^2, describe the transformation from f to g and give the vertex of g.
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Answer. Shift right 3 units and up 2 units; vertex (3, 2)
Replacing x with (x-3) shifts the graph right 3, and adding 2 outside shifts it up 2. Since f's vertex is (0,0), g's vertex becomes (3,2). -
Given f(x) = x^2 + 1 for x < 0 and f(x) = 2x - 3 for x ≥ 0, find f(-2) and f(3).
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Answer. f(-2) = 5 and f(3) = 3
Since -2 < 0, use x^2+1: (-2)^2+1 = 4+1 = 5. Since 3 ≥ 0, use 2x-3: 2(3)-3 = 6-3 = 3. -
Find the domain of f(x) = √(2x - 4) + 1.
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Answer. x ≥ 2
The radicand must be non-negative: 2x-4 ≥ 0, so x ≥ 2. The '+1' outside the root does not affect the domain. -
Describe the transformation that produces g(x) = -∛(x+1) - 3 from the parent function f(x) = ∛x.
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Answer. Reflect over the x-axis, shift left 1 unit, and shift down 3 units
The negative sign in front reflects the graph over the x-axis. Writing (x+1) as (x-(-1)) shows a shift left 1, and the -3 outside shifts the graph down 3. -
Solve 3x^2 - 5x - 2 = 0.
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Answer. x = 2 and x = -1/3
Using the quadratic formula: x = (5 ± √(25+24))/6 = (5 ± 7)/6, giving x=12/6=2 or x=-2/6=-1/3. -
A ball's height is modeled by h(t) = -16t^2 + 64t + 5. Find the maximum height and the time it occurs.
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Answer. Maximum height of 69 feet at t = 2 seconds
The axis of symmetry is t=-64/(2(-16))=2. Then h(2) = -16(4)+64(2)+5 = -64+128+5 = 69 feet. -
How many real solutions does x^2 + 4x + 7 = 0 have?
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Answer. No real solutions
The discriminant is b^2-4ac = 16 - 28 = -12, which is negative, so the equation has no real solutions (two complex solutions). -
A rental company charges \$20 plus \$0.25 per mile for up to 100 miles, then \$0.15 per mile for each mile beyond 100. Write a piecewise function C(m) for the cost and find the cost of a 150-mile trip.
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Answer. C(m) = 20 + 0.25m for 0 ≤ m ≤ 100, and C(m) = 45 + 0.15(m-100) for m > 100; C(150) = \$52.50
At exactly 100 miles the flat-rate piece gives 20+0.25(100)=45, which becomes the starting cost for the second piece. For m=150: 45 + 0.15(50) = 45+7.5 = 52.50. -
Given y = 2√(x-3) - 1, find the domain and range.
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Answer. Domain: x ≥ 3, Range: y ≥ -1
The radicand x-3 must be ≥0, giving domain x≥3. Since √(x-3) ≥ 0, multiplying by 2 keeps it ≥0, and subtracting 1 shifts the minimum output to -1, giving range y≥-1.
What people get wrong
- Dropping the negative sign in x=-b/(2a) — always write the formula with the negative explicitly before substituting values, since a common error is computing b/(2a) instead.
- Getting the sign of h backwards in vertex form — in f(x)=a(x-h)^2+k, a term like (x+3)^2 means h=-3 (shift left 3), not h=3; rewrite (x+3) as (x-(-3)) before reading off h.
- Assuming a square-root function has domain 'all real numbers' — always set the radicand ≥0 and solve for x before stating the domain.
- Misreading which piece of a piecewise function applies at a boundary point — check whether the inequality on each piece is strict or non-strict, since the point can only belong to one piece.
- Declaring a data set 'quadratic' just because the y-values are increasing faster and faster, without actually computing first and second differences to confirm they're constant.
- Forgetting that a negative leading coefficient in a quadratic reflects the parabola (opens downward) when describing transformations, leading to a max instead of a min or vice versa.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of Algebra I. Free, and no account needed to start.