Finding the vertex of \(\displaystyle{y}={x}^{{{2}}}-{6}{x}+{1}\). My solution

Leonard Montes

Leonard Montes

Answered question

2022-04-07

Finding the vertex of y=x26x+1. My solution doesn't match the book.

Answer & Explanation

betazpvaf4

betazpvaf4

Beginner2022-04-08Added 9 answers

Step 1
Any general parabola can be written as L12=4AL2, if L1 and L2 are Eq. of non-parallel lines.
If L1 and L2 are perpendicular and L1 and L2 are normalized, life is simple. The length of latus rectum (LR) is 4A. The Eq. of axis of the parabola is L1=0 Tangen at vertex is L1=0. Vertex is found by solving L1=0 and L2=0. The Eq. of directrix is L2=A, Eq. of LR is L2=A, the focus F come by solving L1=0 & L2=A.
So in your case y=x26x+1y+8=(x3)2 Then L1=x3, L2=y+8, A=14 The vertex is given by L1=0, L2=0 which is x=3, y=8. the vertex is (3,-8)

ti2n1i2mt0l7

ti2n1i2mt0l7

Beginner2022-04-09Added 10 answers

Step 1
Ah, well, both you and the book got the same x coordinate for the vertex: x=3. So let's substitute x=3 into y=x26x+1 and see the y coordinate. It comes out as y=8, so your solution is right.
(This shows there isn't much value in remembering the formula y=4acb24a - you can always just remember the formula for x and then substitute x into the original equality y=ax2+bx+c - the result will be the same.)

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