Rewrite quadratic and exponential functions in equivalent forms that reveal zeros, extreme values, symmetry and percent rate of change

Learning goal: Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. For a quadratic function, use the process of factoring and completing the square to show the zeros, the extreme value and the symmetry of the graph, and interpret these in terms of a context. For an exponential function, use the properties of exponents to interpret the expression, identifying the percent rate of change and classifying the function as exponential growth or decay.

Equivalent forms are equal everywhere and useful differently. Two expressions are equivalent when they give the same output for every input. Rewriting a function into an equivalent form changes nothing about the function and everything about what you can read from it at a glance. A quadratic function has three common forms. The standard form ax² + bx + c shows the vertical intercept c. The factored form a(x − r)(x − s), available when the quadratic has real zeros, shows the zeros r and s. The vertex form a(x − h)² + k shows the vertex (h, k), which is the extreme value k, and the axis of symmetry x = h. Each form hides what the others show, so choosing a form is choosing what to reveal.

A fully worked quadratic model. A hypothetical small theater finds that if it raises its ticket price by x dollars above the current price, the nightly revenue in dollars above a baseline is R(x) = −2x² + 12x + 32. Standard form shows R(0) = 32: with no price change the revenue is 32 dollars above baseline. Factor: −2x² + 12x + 32 = −2(x² − 6x − 16) = −2(x − 8)(x + 2). The zeros are x = 8 and x = −2, so the revenue is at the baseline for a price rise of 8 dollars or a price cut of 2 dollars, and R(x) is positive between those values. Complete the square: x² − 6x − 16 = (x² − 6x + 9) − 9 − 16 = (x − 3)² − 25, so R(x) = −2[(x − 3)² − 25] = −2(x − 3)² + 50. The vertex is (3, 50): the maximum revenue above baseline is 50 dollars, achieved with a 3 dollar price rise, and the axis of symmetry is x = 3. The symmetry means a price rise of 3 + d dollars and one of 3 − d dollars give equal revenue, for example R(1) = R(5) = 42 and R(0) = R(6) = 32. Check the vertex form: at x = 3 the square is zero and R = 50; at x = 0, −2 · 9 + 50 = 32, which matches. Check the factored form: at x = 0, −2 · (−8) · 2 = 32. Three forms, one function, three properties read directly.

Completing the square carefully. The process takes ax² + bx + c, factors a from the first two terms, adds and subtracts the square of half the remaining x-coefficient inside the bracket, and regroups. For 3x² + 12x + 7: 3(x² + 4x) + 7 = 3(x² + 4x + 4 − 4) + 7 = 3(x + 2)² − 12 + 7 = 3(x + 2)² − 5. Vertex (−2, −5), minimum −5, axis x = −2. The step students skip is multiplying the subtracted 4 by the factor 3 before moving it outside: the bracket contains −4, and 3 times −4 is −12, not −4. A quick check restores confidence: at x = 0 the original gives 7 and the vertex form gives 3 · 4 − 5 = 7.

Starting with a leading coefficient of 1. In Algebra 1 and Math 2 the focus is on completing the square in quadratic functions with the leading coefficient of 1, where there is nothing to factor out first. For f(x) = x² + 8x + 7, half of 8 is 4 and 4² = 16, so x² + 8x + 7 = (x² + 8x + 16) − 16 + 7 = (x + 4)² − 9. The vertex is (−4, −9), the minimum is −9 and the axis of symmetry is x = −4. Setting (x + 4)² − 9 = 0 gives x + 4 = ±3, so the zeros are x = −1 and x = −7, which the factored form (x + 1)(x + 7) confirms. Check at x = 0: 7 from the original and 16 − 9 = 7 from the vertex form. A leading coefficient other than 1, as in 3x² + 12x + 7 above, adds the factoring-out step and belongs to Algebra 2 and Math 3.

Which form for which question. Where does the profit vanish? Factored form. What is the greatest height and when? Vertex form. What was the starting value? Standard form. When factoring is not reasonable, because the zeros are not rational, completing the square still works and the zeros can then be read by solving a(x − h)² + k = 0 for x; the quadratic formula is completing the square done once in general.

Exponential functions and the properties of exponents. An exponential function y = a · bᵗ, with a positive starting value a, has growth factor b per unit of t. When b > 1 the function represents exponential growth and the percent rate of change is (b − 1) × 100 percent per unit; when 0 < b < 1 it represents exponential decay and the percent rate of change is (1 − b) × 100 percent per unit. So y = (1.02)ᵗ grows by 2 percent per unit of t, and y = (0.97)ᵗ decays by 3 percent per unit. In Algebra 1 and Math 2 the focus is on exponential functions evaluated at integer inputs, where the properties of exponents can be seen in a table. For y = 500(1.04)ᵗ the values at t = 0, 1, 2 and 3 are 500, 520, 540.80 and about 562.43: each is 1.04 times the one before, because 500(1.04)ᵗ⁺¹ = 500(1.04)ᵗ · 1.04, so the quantity grows 4 percent per step and 500 is the value at t = 0. For y = 800(0.75)ᵗ the values are 800, 600, 450 and 337.5: each keeps 75 percent of the one before, a decay of 25 percent per step. The rewritings below use exponents that are not integers, such as m/12 and h/3; they go beyond that focus and belong to Algebra 2 and Math 3. The properties of exponents let you read the rate for a different unit of time. Suppose an account is modeled by A(t) = 1000(1.08)ᵗ with t in years, and you want the monthly rate. Let m = 12t be the number of months, so t = m/12 and A = 1000(1.08)^(m/12) = 1000[(1.08)^(1/12)]ᵐ, using (bᵖ)ᵠ = bᵖᵠ. Computing, (1.08)^(1/12) ≈ 1.00643, so the balance grows by about 0.643 percent per month. Going the other way, a quantity growing 2 percent per month is y = (1.02)ᵐ = (1.02)^(12t) = [(1.02)¹²]ᵗ ≈ (1.268)ᵗ with t in years, a yearly growth of about 26.8 percent, not 24 percent, because the monthly growth compounds. The same property rewrites 2^(3t) as 8ᵗ, and a half-life model 100 · (1/2)^(t/5) as 100 · [(1/2)^(1/5)]ᵗ ≈ 100(0.8706)ᵗ, revealing a decay of about 12.9 percent per unit of t.

The misconception about decay. A function y = 5(0.97)ᵗ is sometimes read as decreasing by 97 percent, or as growing because 0.97 is positive. The factor 0.97 means that each step keeps 97 percent of the previous amount and loses 3 percent. Growth or decay is decided by whether the base is above or below 1, never by its sign, and the percent change is the distance of the base from 1. Similarly (1.5)ᵗ is a 50 percent increase per unit, not 150 percent.

Worked factoring and completing the square: A ball's height in meters is h(t) = −5t² + 20t + 25. Factored: −5(t² − 4t − 5) = −5(t − 5)(t + 1), zeros t = 5 and t = −1; the ball lands at t = 5 and t = −1 is outside the flight. Completing the square: t² − 4t − 5 = (t − 2)² − 4 − 5 = (t − 2)² − 9, so h(t) = −5(t − 2)² + 45. Vertex (2, 45): the maximum height is 45 meters at 2 seconds, and the parabola is symmetric about t = 2, so the ball is at the same height 1 second before and 1 second after its peak. Check: h(0) from vertex form is −5 · 4 + 45 = 25, matching the standard form.

Worked exponential interpretation: A population is modeled by P(t) = 3000(0.94)ᵗ with t in years. The base 0.94 is less than 1, so this is exponential decay, at 1 − 0.94 = 0.06, a 6 percent decrease per year. To read the decade rate, write t = 10d with d in decades: P = 3000(0.94)^(10d) = 3000[(0.94)¹⁰]ᵈ, and (0.94)¹⁰ ≈ 0.5386, so the population loses about 46.1 percent per decade. It is not 60 percent, because each year's 6 percent is taken from a smaller amount than the year before.

Worked change of time unit: A culture grows according to N(h) = 50 · 2^(h/3) with h in hours, doubling every 3 hours. Rewrite with the property (bᵖ)ᵠ = bᵖᵠ: 2^(h/3) = (2^(1/3))ʰ ≈ (1.2599)ʰ, so the hourly growth rate is about 26.0 percent. Alternatively, in units of 3-hour periods p, N = 50 · 2ᵖ, growth of 100 percent per period. All three expressions are equivalent and describe the same culture; the base changes with the unit of time while the function does not.

Representation task: For R(x) = −2x² + 12x + 32, write the three forms side by side, standard, factored and vertex, and under each write the property it reveals and the value of that property. Make a table of R for x = −2 through 8 and mark the zeros, the vertex and two symmetric pairs, then sketch the graph and label all of them. Verify each form by substituting x = 0 and x = 1 into all three and showing the outputs agree.

Independent performance: Write a quadratic function that models a situation of your own, with a maximum or minimum that means something, and convert it to factored and vertex form, interpreting the zeros, the extreme value and the axis of symmetry in the context. Then write two exponential functions, one growth and one decay, given per year, and use the properties of exponents to rewrite each with a monthly base, stating the percent rate of change for both units. Success criteria: every form is checked by substitution at two inputs; the vertex is confirmed by completing the square with the coefficient handled correctly; each zero is interpreted as a statement about the situation; the growth or decay classification is justified by comparing the base with 1; and the monthly rate is computed as a twelfth root rather than by dividing the yearly percent by 12.