# 2. Find the point on the curve y= x2 – 6x + 3 where the tangent is horizontal?

### 1 Answer

- PuzzlingLv 73 months ago
ALGEBRAIC METHOD:

You have a quadratic of the form:

y = ax² + bx + c

That function represents a parabola. The tangent will be horizontal at the vertex. And the x-coordinate of the vertex can be found using this formula:

x = -b/(2a)

(If you need help remembering that, it's essentially the quadratic formula but without the ±√ part)

Plug in your values:

x = -(-6) / (2*1)

x = 6/2

x = 3

Finally calculate the y-value by plugging x=3 to the original function:

y = 3² - 6(3) + 3

y = 9 - 18 + 3

y = -6

Answer:

(3, -6)

CALCULUS METHOD:

Original function:

y = x² - 6x + 3

The derivative represents the slope at each point, so take the derivative:

y' = 2x - 6

You want the tangent to be horizontal (slope = 0) so set that to 0.

2x - 6 = 0

2x = 6

x = 6/2

x = 3

That gets you the x-coordinate. Plug in x=3 into the original equation to get the y-coordinate:

y = 3² - 6(3) + 3

y = 9 - 18 + 3

y = -6

Answer:

(3, -6)