Graphing and solving quadratics on the HP Prime — vertex/standard/factored forms, complex numbers, completing the square, the quadratic formula, and linear-quadratic systems.
Problem types: Graphing y = a(x − h)² + k; identifying the vertex (h, k), axis of symmetry, min/max, domain, and range.
🎣 Method (teach-the-tool)
Tool
Function app (Symbolic + Plot, Fcn ▸ Extremum)
Why this tool
Vertex form is built for graphing. Graph it in the Function app and let Extremum read the vertex straight off the curve — no algebra needed to confirm (h, k).
Path
Apps ▸ Function ▸ Symb ▸ highlight F1(X) ▸ enter 2*(X-3)^2+4 ▸ Enter ▸ Plot ▸ trace near the bottom of the curve ▸ tap Menu ▸ tap Fcn ▸ select Extremum
📝 Worked example
Problem: Graph f(x) = 2(x − 3)² + 4 and identify the vertex, axis of symmetry, min/max, domain, and range.
Do this on the calculator:
AppsPress the Apps key to open the App Library — the screen with all the app icons.
FunctionHighlight the Function app and press Enter to open it. It starts on the Symb (symbolic/equation) screen, where you type the function.
Type2*(X-3)^2+4With F1(X) highlighted, type the function. Use the [X,θ,T,N] key for X and the x^y key for the power.
EnterLocks in the definition for F1(X).
PlotPress the Plot key to draw the graph — you'll see a U-shaped parabola.
trace near the bottom of the curveUse the arrow keys to move the blinking tracer close to the lowest point of the U.
Menu▸Fcn▸ExtremumTap the Menu soft button, then tap Fcn, then pick Extremum. It finds the turning point (the peak or valley) and prints its coordinates at the bottom.
You'll see: The graph's lowest point is highlighted and the status line shows Extremum: (3, 4) — that's the vertex (h, k). Since a = 2 > 0 it opens up, so it's a minimum: axis of symmetry x = 3, range [4, ∞), domain (−∞, ∞).
✅ Quick-verify (check your hand-work)
read the Extremum coordinatesLook at the (3, 4) the calculator just printed at the bottom of the Plot screen.
compare to vertex formIn y = a(x − h)² + k the vertex is (h, k) = (3, 4). The calculator's answer matches, so you read the form correctly.
Problem types: Finding the vertex and axis of symmetry of y = ax² + bx + c; y-intercept (0, c); graphing from standard form.
🎣 Method (teach-the-tool)
Tool
Home view ( −b/2a ) with Function app (Fcn ▸ Extremum) to confirm
Why this tool
In standard form the vertex x-coordinate is h = −b/2a — a quick Home calculation. Plug h back in for k, then graph and use Extremum to confirm. The y-intercept is just c, read at x = 0.
Path
Home ▸ -(-6)/(2*1) ▸ Enter (gives h) ▸ then Apps ▸ Function ▸ Symb ▸ F1(X)=X^2-6*X+10 ▸ Plot ▸ Menu ▸ Fcn ▸ Extremum
📝 Worked example
Problem: Find the vertex, axis of symmetry, and y-intercept of f(x) = x² − 6x + 10.
Do this on the calculator:
HomePress the Home key for plain number-crunching.
Type-(-6)/(2*1)Type the vertex x-formula −b/2a with a = 1, b = −6. Use the (−) negative-sign key (the small one, not the minus key) for the leading minus.
EnterThe screen prints 3 — that's h, the x-coordinate of the vertex.
AppsPress the Apps key to open the App Library.
FunctionHighlight the Function app and press Enter to open it, landing on the Symb (equation) screen.
TypeX^2-6*X+10With F1(X) highlighted, type the function. Use the [X,θ,T,N] key for X and the x^y key for the square.
EnterLocks in the definition for F1(X).
PlotPress the Plot key to draw the parabola.
trace near the turning pointUse the arrow keys to move the tracer close to the bottom of the curve.
Menu▸Fcn▸ExtremumTap Menu, then Fcn, then Extremum. It prints the vertex coordinates at the bottom.
You'll see: Home shows 3 — that's h. The graph's lowest point is highlighted and the status line shows Extremum: (3, 1) — the vertex. Axis of symmetry x = 3; y-intercept is (0, c) = (0, 10), confirmable in Num view at the X=0 row.
✅ Quick-verify (check your hand-work)
HomeSwitch to the Home screen.
Type-(-6)/(2*1)Re-type the −b/2a formula for the vertex x.
EnterPrints 3 — the same x as the Extremum vertex, so h is confirmed.
NumBack in the Function app, press the Num key for the table, then scroll to the X = 0 row: F1 reads 10 — the y-intercept c.
Problem types: Writing a quadratic in factored form; using the Zero Product Property to find zeros / x-intercepts; solving quadratic equations by factoring.
🎣 Method (teach-the-tool)
Tool
CAS factor() for the factored form; Function app (Fcn ▸ Root) for the x-intercepts
Why this tool
factor() turns standard form into the factored form whose zeros you read off directly, and Fcn ▸ Root finds each x-intercept on the graph — the two views of the same zeros.
Path
CAS ▸ Toolbox ▸ tap CAS ▸ choose factor (or type factor) ▸ factor(x^2+2*x-8) ▸ Enter | graph: Apps ▸ Function ▸ Symb ▸ F1(X)=X^2+2*X-8 ▸ Plot ▸ trace near an x-intercept ▸ Menu ▸ Fcn ▸ Root (repeat near the other intercept)
📝 Worked example
Problem: Write y = x² + 2x − 8 in factored form and find its zeros (x-intercepts).
Do this on the calculator:
CASPress the CAS key to open the CAS (symbolic algebra) screen, where the calculator can factor.
Typefactor(x^2+2*x-8)Type the factor command around your quadratic. Type the command name with the letter keys (or open Toolbox ▸ tap CAS ▸ pick factor). Use the x^y key for the square.
EnterCAS returns the factored form (x − 2)·(x + 4).
You'll see: CAS returns (x − 2)·(x + 4). By the Zero Product Property the zeros are x = 2 and x = −4 — the x-intercepts (2, 0) and (−4, 0). Confirm on the graph with Fcn ▸ Root near each crossing: Root = 2, then Root = −4.
✅ Quick-verify (check your hand-work)
Apps▸FunctionOpen the App Library with Apps, highlight Function, press Enter, and on the Symb screen enter F1(X)=X^2+2*X-8.
PlotPress the Plot key to draw the parabola, which crosses the x-axis twice.
trace near one crossingUse the arrow keys to move the tracer close to one x-intercept.
Menu▸Fcn▸RootTap Menu, then Fcn, then Root. It prints the nearest x-intercept — here 2.
trace to the other crossing, then Menu▸Fcn▸Root againRoot finds only the root nearest the tracer, so move to the other crossing and repeat — it prints −4. Both match the factors' zeros.
Problem types: Adding, subtracting, and multiplying complex numbers; simplifying a quotient to a + bi form; using i² = −1; solving x² = a when a < 0.
🎣 Method (teach-the-tool)
Tool
Home view complex arithmetic ( a + b*i ); CAS cSolve for complex roots
Why this tool
The Home view evaluates complex expressions directly in a + bi form — type the expression with i and press Enter. For equations with non-real roots, cSolve returns the complex solutions.
Path
Home ▸ enter the expression with i, e.g. (3-2*i)*(3+2*i) ▸ Enter | complex roots: CAS ▸ cSolve(x^2+5=0, x) ▸ Enter
📝 Worked example
Problem: Multiply the conjugates (3 − 2i)(3 + 2i), then simplify the quotient 10/(2 − i) to a + bi form.
Do this on the calculator:
HomePress the Home key for plain calculation.
Type(3-2*i)*(3+2*i)Type the product. To get the imaginary i, press the green Shift key then the key printed i. Use the (−) negative-sign key inside, not the minus key, only for negative numbers — here the − between terms is the regular minus key.
EnterThe screen prints 13 — the i terms cancel, so conjugates give a real result.
Type10/(2-i)Type the quotient. Again use green Shift then the i key for the imaginary unit.
EnterThe screen prints 4 + 2*i — the quotient written in a + bi form.
You'll see: The Home screen prints 13 for the first (the i terms cancel — a real result). The second prints 4 + 2*i, the quotient in a + bi form.
✅ Quick-verify (check your hand-work)
HomeSwitch to the Home screen.
Type(3-2*i)*(3+2*i)Re-type your hand result's expression (green Shift then the i key for i).
EnterPrints 13 — matches your worked answer.
Type10/(2-i) ▸ EnterType the quotient and press Enter — it prints 4 + 2i, confirming the a + bi form.
Problem types: Transforming x² + bx + c = 0 into (x − p)² = q; writing a quadratic in vertex form by completing the square; revealing the min/max value.
🎣 Method (teach-the-tool)
Tool
CAS expand() to confirm the completed form; Function app (Fcn ▸ Extremum) to confirm the vertex
Why this tool
Completing the square by hand is the skill — the calculator just checks it. expand() of your completed/vertex form should reproduce the original standard form, and Extremum confirms the vertex you revealed.
Path
CAS ▸ expand((x+4)^2-11) ▸ Enter (should reproduce the original) | vertex check: Apps ▸ Function ▸ Symb ▸ F1(X)=-2*X^2+10*X+1 ▸ Plot ▸ Menu ▸ Fcn ▸ Extremum
📝 Worked example
Problem: Solve x² + 8x + 5 = 0 by completing the square, and check the completed form with the calculator.
Do this on the calculator:
by hand: (x + 4)² = 11Complete the square yourself first — this is the skill being graded. It gives x = −4 ± √11. The calculator only checks it.
CASPress the CAS key to open the symbolic algebra screen.
Typeexpand((x+4)^2-11)Type the expand command around your completed form (x + 4)² − 11. Use the x^y key for the square.
EnterCAS multiplies it out back to x² + 8x + 5 — the original, so your completed square is correct.
You'll see: CAS expands to the original standard form x² + 8x + 5 (the calculator may order the terms differently, e.g. x^2+8*x+5) — confirming the completed square (x + 4)² − 11 is equivalent. The roots are x = −4 ± √11.
✅ Quick-verify (check your hand-work)
CASOpen the CAS screen.
Typeexpand((x+4)^2-11) ▸ EnterExpand your completed form — it must return x² + 8x + 5, the original standard form.
vertex-form check (optional)For the −2x² + 10x + 1 example, graph it in the Function app and run Menu ▸ Fcn ▸ Extremum — it returns (2.5, 13.5), matching the completed vertex.
Problem types: Solving ax² + bx + c = 0 with the quadratic formula (real AND complex roots); using the discriminant b² − 4ac to predict the number/type of roots.
🎣 Method (teach-the-tool)
Tool
CAS solve() / proot([a,b,c]); Solve app for a numeric solution
Why this tool
solve() returns the exact roots — real or complex — matching the formula, and proot takes just the coefficient list [a, b, c]. Use these to confirm the roots you found by hand.
Path
CAS ▸ solve(3*x^2-4*x-9=0, x) ▸ Enter | or proot([3,-4,-9]) ▸ Enter | complex case: solve(x^2-9*x+27=0, x)
📝 Worked example
Problem: Solve 3x² − 4x − 9 = 0 with the quadratic formula and confirm on the calculator.
Do this on the calculator:
CASPress the CAS key to open the symbolic algebra screen.
Typesolve(3*x^2-4*x-9=0, x)Type the solve command with your equation and the variable x. Use the x^y key for the square; type the command name with the letter keys.
EnterCAS returns the two exact roots as a set, matching the quadratic formula.
Typeproot([3,-4,-9])For decimal values, type proot with just the coefficient list [a, b, c] = [3, −4, −9]. Use the (−) negative-sign key inside the brackets.
EnterReturns the same two roots as a decimal vector.
You'll see: solve returns a two-element set holding the exact roots (2 + √31)/3 and (2 − √31)/3 (matching the formula's (4 ± √124)/6). proot([3,-4,-9]) returns the same two roots as a decimal vector. The exact surd layout on screen may differ from the printed form — confirm the values, not the typography.
✅ Quick-verify (check your hand-work)
HomePress the Home key for plain calculation.
Typethe expression with a root plugged inSubstitute one solved root back into 3*X^2-4*X-9 (type the number in place of X). A correct real root makes the expression evaluate to about 0.
EnterThe screen shows ~0 (a tiny rounding value is fine), confirming the root.
Problem types: Finding the number of solutions to a line/parabola system (0, 1, or 2); solving a linear-quadratic system by graphing or substitution; using a system to solve an equation.
🎣 Method (teach-the-tool)
Tool
Function app — graph both, then Fcn ▸ Intersection; CAS solve() for the system
Why this tool
Graph the line and the parabola together and Fcn ▸ Intersection reads each crossing — the solution(s). CAS solve() of the system confirms the same points algebraically.
Path
Apps ▸ Function ▸ Symb ▸ F1(X)=2*X^2-6*X-8 ▸ F2(X)=2*X-16 ▸ Plot ▸ trace near a crossing ▸ Menu ▸ Fcn ▸ Intersection ▸ pick the other function | CAS ▸ solve({y=2*x^2-6*x-8, 2*x-y=16}, {x,y})
📝 Worked example
Problem: Solve the system y = 2x² − 6x − 8 and 2x − y = 16 by graphing.
Do this on the calculator:
rewrite the line as y = 2x − 16On paper, solve 2x − y = 16 for y so you can enter it as a function.
AppsPress the Apps key to open the App Library.
FunctionHighlight the Function app and press Enter to open it on the Symb screen.
Type2*X^2-6*X-8 into F1(X)With F1(X) highlighted, type the parabola. Use the [X,θ,T,N] key for X and the x^y key for the square; press Enter.
Type2*X-16 into F2(X)Move down to F2(X) and type the line, then press Enter.
PlotPress the Plot key to draw both the parabola and the line together.
trace near a crossingUse the arrow keys to move the tracer close to where the two curves meet.
Menu▸Fcn▸IntersectionTap Menu, then Fcn, then Intersection. It asks which other curve to use.
pick F2 (the line)Choose F2 from the list. The calculator prints the crossing point at the bottom.
You'll see: The crossing is marked and the status line shows Intersection: (2, −12) — the solution. (Intersection returns only the crossing nearest the cursor; here the parabola and line meet at exactly one point.)
✅ Quick-verify (check your hand-work)
HomePress the Home key for plain calculation.
Type2*(2)^2-6*(2)-8Plug the x-value of the crossing into the parabola to check it.
EnterPrints −12 — the y-value of the crossing, so the point is on the parabola.
Type2*(2)-(-12) ▸ EnterPlug the point into the line 2x − y. It prints 16 — matches, so the point satisfies BOTH equations. (Intersection returns only the crossing nearest the cursor — move the cursor and run it again to find another.)