How do you find the vertex and the intercepts for $y=(x+3)(x-1)$ ?

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Answer 1 · 2016-08-30T19:12:53

Vertex: $(-1, -4)$
x-intercepts: $(-3, 0)$ and $(1, 0)$
y-intercept: $(0, -3)$

Since the function is already factored, we can note the x-intercepts.

$0 = (x+ 3)(x - 1)$

$x = -3 and 1$

Hence, the x-intercepts are at $(-3, 0)$ and $(1, 0)$ .

We should now multiply out.

$y = (x + 3)(x - 1)$

$y =x^2 + 3x - x - 3$

$y = x^2 + 2x - 3$

We can now find the y-intercept very easily.

$y = 0^2 + 2(0) - 3$

$y = -3$

Hence, the y-intercept is at $(0, -3)$ .

As for the vertex, it'll be easiest to complete the square to convert into vertex form, $y = a(x - p)^2 + q$ .

$y = x^2 + 2x - 3$

$y = 1(x^2 + 2x + n - n) - 3$

$n = (b/2)^2 = (2/2)^2 = 1$

$y = 1(x^2 + 2x + 1 - 1) - 3$

$y= 1(x^2 + 2x + 1) - 1 - 3$

$y = 1(x + 1)^2 - 4$

The vertex in vertex form $y = a(x - p)^2 + q$ is located at $(p, q)$ . Thus, our vertex is at $(-1, -4)$ .

Hopefully this helps!

How do you find the vertex and the intercepts for y=(x+3)(x-1)y=(x+3)(x-1)?