See it in action

1. Meet the problem

What is the vertex of this parabola? $y={\left(x-3\right)}^2+2$

2. A common sign mistake

What is the vertex of this parabola? $y={\left(x-3\right)}^2+2$

Student: (-3, 2) · Not quite

3. Sage helps the student reason

What is the vertex of this parabola? $y={\left(x-3\right)}^2+2$

Student: (-3, 2) · Not quite

Tutor: You found the vertical coordinate, 2. Let’s check the horizontal one together. The square is smallest when x − 3 equals zero. Which x makes that happen?

$x-3=0$

4. Connect it to a Principle

What is the vertex of this parabola? $y={\left(x-3\right)}^2+2$

Student: (-3, 2) · Not quite

Tutor: That happens at x = 3. Then the square is zero and y = 2. The minus inside the parentheses means a shift to the right. You can always check by substitution.

$y={\left(3-3\right)}^2+2y=(3-3{)}^2+2=2$

5. Try again, with understanding

What is the vertex of this parabola? $y={\left(x-3\right)}^2+2$

Student: (3, 2) · Correct

Tutor: Yes! The vertex is (3, 2). You corrected the direction of the shift and checked why it works.

6. A fresh follow-up problem

Try the same idea with a different shift. Find the vertex. $y={\left(x+4\right)}^2-1$

Tutor: Let’s use that idea again. What value makes x + 4 equal zero?

7. Correct on the first try

Try the same idea with a different shift. Find the vertex. $y={\left(x+4\right)}^2-1$

Student: (-4, -1) · Correct

Tutor: Exactly. At x = −4 the square is zero, leaving y = −1. You used the Principle on a new problem.

$y={\left(-4+4\right)}^2-1y=(-4+4{)}^2-1=-1$

1. Meet the problem

Solve the quadratic. Find every value of x. $x^2-5x+6=0$

2. A correct first answer

Solve the quadratic. Find every value of x. $x^2-5x+6=0$

Student: x = 2 or x = 3 · Correct

3. Scout checks the reasoning

Solve the quadratic. Find every value of x. $x^2-5x+6=0$

Student: x = 2 or x = 3 · Correct

Tutor: Both answers check out. Here’s the clue: which two numbers multiply to 6 and add to −5? They are −2 and −3, so we can factor the quadratic.

$x^2-5x+6=\left(x-2\right)\left(x-3\right)$

4. Know why both roots count

Solve the quadratic. Find every value of x. $x^2-5x+6=0$

Student: x = 2 or x = 3 · Correct

Tutor: Either factor can make the product zero. Setting x − 2 = 0 gives 2; setting x − 3 = 0 gives 3. That explains both solutions.

5. A follow-up catches a missing root

One more quadratic. Remember to find every solution. $x^2+x-6=0$

Student: x = 2 · Not quite

Tutor: You found a valid solution. Is it the only one? Let’s investigate the other factor.

6. Follow the second clue

One more quadratic. Remember to find every solution. $x^2+x-6=0$

Student: x = 2 · Not quite

Tutor: The factors are x + 3 and x − 2. You used the second factor. Setting the first one equal to zero gives x = −3, another solution.

$x^2+x-6=\left(x+3\right)\left(x-2\right)$

7. The complete answer

One more quadratic. Remember to find every solution. $x^2+x-6=0$

Student: x = -3 or x = 2 · Correct

Tutor: Case closed: −3 and 2 both make the original expression zero. Check every factor so no solution gets left behind.

${\left(-3\right)}^2+\left(-3\right)-6=0$