Question

The coordinates of the turning point of the parabola whose equation is y = x2 — 6x + 8 are A. (3, 35) C. (-3, 35) B. (-3, —1) D. (3. —1)

Answer 1

Answer:

D. (3. —1)

Step-by-step explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

$f(x) = ax^{2} + bx + c$

It's vertex is the point $(x_{v}, y_{v})$

In which

$x_{v} = -\frac{b}{2a}$

$y_{v} = -\frac{\Delta}{4a}$

Where

$\Delta = b^2-4ac$

If a<0, the vertex is a maximum point, that is, the maximum value happens at $x_{v}$, and it's value is $y_{v}$.

Turning point of a quadratic function:

The turning point of a quadratic function is the vertex.

y = x2 — 6x + 8

This means that $a = 1, b = -6, c = 8$

Now, we find the vertex.

$x_{v} = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3$

$\Delta = b^2-4ac = (-6)^2-4(1)(8) = 36 - 32 = 4$

$y_{v} = -\frac{4}{4(-1)} = -1$

(x,y) = (3,-1), and the correct answer is given by option D.