Question

What are the vertex and x-intercepts of the graph of the function given below?

y = x^2 - 2x - 35

A. Vertex: (0, 0); intercepts: x = -4, -5
B. Vertex: (1, -36); intercepts: x = 7, -5
C. Vertex: (1, -25); intercepts: x = 6, -4
D. Vertex: (7, 5); intercepts: x = 7, 8

Answer 1

@Jaynator495 @zepdrix

Answer 2

hint:
The vertex of a parabola
y=ax^2+bx+c
is located at x=-b/a
therefore the coordinates of the vertex is (-b/a, y(-b/a) )

Answer 3

Do-what?

Answer 4

@ShadowLegendX

Answer 5

To find the vertex, you need to complete the square of the given function. Have you any experience with completing the square?

Answer 6

y = x^2 - 2x - 35 can be rewritten as

y = x^2 - 2x - 35

Take HALF of the coefficient of x (which is -2); square that and add it to the first two terms, and then subtract the same thing.

OK I undrstand you need to go. Come back later.

Answer 7

Once again: Pls apply all you know about completing the square to finding the vertex here. Do y ou have a printed textbook that might discuss this topic?

Answer 8

The x-coordinate of the vertex is
\[
-\frac b{2a}
\]

Answer 9

It is not as mentioned above as -b/a

Answer 10

\[
x^2-2x-35=x^2-2x+1-1-35=(x^2-2x +1) -36\]
Can you continue from here?

Answer 11

@eliesaab
Thank you for the good catch! Yes, indeed, the coordinates of the vertex should be
(-b/(2a), y(-b/(2a) )
However, @mathmale 's suggestion of solving it by completing the square is even better, because there will be nothing to memorize. We see how memory is unreliable! lol