Question

Select all of the ordered pairs that make the inequality y > x2 + 3x – 4 true.

Answer 1

Answer:

$\boxed{x\in(-\infty,\ -4)\ \cup\ (1,\ \infty)}\\or\\\boxed{x<-4\ \vee\ x>1}$

Step-by-step explanation:

$y>x^2+3x-4$

STEP 1:

Find the zeros of a parabola.

$x^2+3x-4=0\\\\x^2+4x-x-4=0\\\\x(x+4)-1(x+4)=0\\\\(x+4)(x-1)=0\iff x+4=0\ or\ x-1=0\\\\x+4=0\qquad\text{subtract 4 from both sides}\\x=-4\\\\x-1=0\qquad\text{add 1 to both sides}\\x=1$

STEP 2:

Sketch a parabola.

The coefficient at x² is 1. It's a positive number, therefore the parabola is open up.

(attachment #1)

STEP 3:

Mark the regions where the graph is above the axis.

(attachment #2)

STEP 4:

Read selected intervals

$x\in(-\infty,\ -4)\ \cup\ (1,\ \infty)$

or

$x<-4\ \vee\ x>1$

Answer 2

Answer:

A.) (0,0)

B.) (-2,-1)

Answer on edge 2020

Step-by-step explanation: