Question

~~~~~~~~~~~FAN AND MEDAL!~~~~~~~~~~~~~~~~
What are the x-intercept(s) of the graph of y − 3 = x2 + 4x?

(−3, 0) and (−1, 0)

(3, 0) and (1, 0)

(−3, 0) and (1, 0)

(3, 0) and (−1, 0)

Answer 1

It's just a somewhat convoluted way to ask you to solve the quadratic equation. The x-intercept is where y = 0 so you're just solving \(0 - 3 = x^2 + 4x\)
In other words
\[x^2 - 4x + 3 = 0\]
Use whatever method you like best.
(I tend to prefer the quadratic formula, but this one will factor nice enough I guess)

Answer 2

Oops, I copied it wrong. \(x^2 + 4x + 3\)

Answer 3

lol its okay, can you show me how we would put my question into that formula?

Answer 4

Ok, the quadratic formula is
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Does that look familiar?

Answer 5

Its coming back to me a bit

Answer 6

The standard form of the equation is \(ax^2 + bx + c\) which is where we get those values. So for \(x^2 + 4x + 3\) it's just:
\[x = \frac{-4 \pm \sqrt{4^2 - 4(1)(3)}}{2(1)}\]
So just crunch that and you'll get your two solutions.

Answer 7

i keep getting negative 2 @SACAPUNTAS

Answer 8

@GirlByte

Answer 9

\[\frac{-4 \pm \sqrt{16 - 12}}{2}\]\[\frac{-4 \pm \sqrt{4}}{2}\]\[\frac{-4 \pm 2}{2}\]\[-2 \pm 1\]
So -1 and -3.

Answer 10

(-3,0) (-1,0)

Answer 11

Yes.