Checking Vertex Calculation

[dxs]

Hello! I have a function, f(x) = x² - x - 6, and have to get the coordinates of the vertex. I want to make sure I'm doing this right.

Coordinates of the Vertex:
Vertex: (½, -25/4)

Step 1: Factor the function:
f(x) = x² - x – 6
f(x) = (x+2)(x-3)

Step 2: Set (x+2) and (x-3) equal to 0 and solve for x:
x + 2 = 0
x = -2

x - 3 = 0
x = 3

The x-intercepts are -2 and 3.

Step 3: To determine the x-value of the vertex, find the average:
[x₁ + x₂] / 2 = [(-2) + (3)] / 2 = ½..........................................[edited]

Step 4: To determine the y-value of the vertex, substitute the x-value of the vertex into the original equation and solve for y:
f(x) = x² - x – 6
f(½) = (½)² – (½) – 6
f(½) = ¼ - ½ - 6
f(½) = -25/4

 

[Deleted member 4993]

Hello! I have a function, f(x) = x² - x - 6, and have to get the coordinates of the vertex. I want to make sure I'm doing this right.

Coordinates of the Vertex:
Vertex: (½, -25/4)

Step 1: Factor the function:
f(x) = x² - x – 6
f(x) = (x+2)(x-3)

Step 2: Set (x+2) and (x-3) equal to 0 and solve for x:
x + 2 = 0
x = -2

x - 3 = 0
x = 3

The x-intercepts are -2 and 3.

Step 3: To determine the x-value of the vertex, find the average:
x₁ + x₂ / 2 = (-2) + (3) / 2 = ½

Step 4: To determine the y-value of the vertex, substitute the x-value of the vertex into the original equation and solve for y:
f(x) = x² - x – 6
f(½) = (½)² – (½) – 6
f(½) = ¼ - ½ - 6
f(½) = -25/4

Thank you for showing your work in an organized way.

As far as I can see, you are CORRECT. Again - nice and neat presentation.

 

[Dr.Peterson]

Hello! I have a function, f(x) = x² - x - 6, and have to get the coordinates of the vertex. I want to make sure I'm doing this right.

Coordinates of the Vertex:
Vertex: (½, -25/4)

Step 1: Factor the function:
f(x) = x² - x – 6
f(x) = (x+2)(x-3)

Step 2: Set (x+2) and (x-3) equal to 0 and solve for x:
x + 2 = 0
x = -2

x - 3 = 0
x = 3

The x-intercepts are -2 and 3.

Step 3: To determine the x-value of the vertex, find the average:
x₁ + x₂ / 2 = (-2) + (3) / 2 = ½

Step 4: To determine the y-value of the vertex, substitute the x-value of the vertex into the original equation and solve for y:
f(x) = x² - x – 6
f(½) = (½)² – (½) – 6
f(½) = ¼ - ½ - 6
f(½) = -25/4

There are, of course, several ways this could have been done; this is not "the right way" (which you didn't say it was), but "a right way" (and perhaps the quickest in this case).

But be careful: What you wrote as x₁ + x₂ / 2 should have been (x₁ + x₂) / 2.

And since this would not have worked if the polynomial could not be factored, make sure you have other methods available.

 

[HallsofIvy]

I have always had a preference for "completing the square".

The function is \(\displaystyle f(x)= x^2- x- 6\). A "complete square" is of the form \(\displaystyle (x- a)^2= x^2- 2ax+ a^2\). Comparing \(\displaystyle x^2- x\) with \(\displaystyle x^2- 2ax\), a must be 1/2. Then \(\displaystyle a^2= 1/4\) so we can "complete the square" by adding 1/4: \(\displaystyle x^2- x+ 1/4= (x- 1/2)^2\). Then \(\displaystyle x^2- x- 6= x^2- x+ 1/4- 1/4- 6= (x- 1/2)^2- 25/4\). Since a square of a real number is never negative, that is least when x= 1/2 where y= -25/4.

 

[Otis]

… f(x) = x² - x - 6 … get the coordinates of the vertex …

Hello dxs. Here's another method, for finding the x-coordinate of the vertex.

Given a quadratic polynomial Ax^2 + Bx + C, the x-coordinate of the vertex is -B/(2A).

I can remember -B/(2A) because that's what the Quadratic Formula yields when the discriminant is zero.

C - B^2/(4A) is the corresponding expression for the y-coordinate, but I can never seem to remember that one, so I evaluate the polynomial at the x-coordinate (just like you did).

?