This is called completing the squares:
y=x²-8x+10
y=(x-4)²-16 + 10
y=(x-4)² - 6
The vertex is thus at (4,-6).
You can also get the vertex directly from:
(-b/2a, -Δ/4a)
=(-(-8)/2, -((-8)²-4*1*10)/4a)
=(4, -6)
y=x^2-8x+10
y=x²-8x+10
y=(x-4)²-16 + 10
y=(x-4)² - 6
The vertex is thus at (4,-6).
You can also get the vertex directly from:
(-b/2a, -Δ/4a)
=(-(-8)/2, -((-8)²-4*1*10)/4a)
=(4, -6)
y = a(x - h)^2 + k
Where (h, k) represents the coordinates of the vertex.
Let's start with your quadratic equation: y = x^2 - 8x + 10
Step 1: Group the x-terms together and leave space to complete the square: y = (x^2 - 8x) + 10
Step 2: Take half of the coefficient of the x-term (-8) and square it to find the constant term to complete the square. In this case, (-8/2)^2 = (-4)^2 = 16.
Step 3: Add the constant term inside the parentheses while subtracting it outside the parentheses to maintain the equation's balance: y = (x^2 - 8x + 16 - 16) + 10
Step 4: Rearrange the equation by combining like terms: y = (x^2 - 8x + 16) - 16 + 10
Simplified: y = (x - 4)^2 - 6
The equation is now in vertex form, where the vertex is represented by (h, k). In this case, the vertex is (4, -6).