If x^2 + 4x + n > 13 for all x, then which of the following must be tr

\(x^2 + 4x + n > 13\)
\((x+2)^2 > 17-n\)
If n=17; x=-2
But if n>17; 17-n <0; (x+2)^2> 17-n for all values of x

IMO A

Concept: the Graph of a Quadratic Expression will produce a Parabola in the coordinate plane in which the Input values (x) will produce corresponding Output values (y) such that the graph will be “U-shaped”

Step 1: change the quadratic expression into Vertex Form

(x)^2 + 4x + n > 13

(x)^2 + 4x + 4 - 4 + n > 13

(x + 2)^2 - 4 + n > 13

(x + 2)^2 - 17 + n > 0

step 2: analyze the Parabola

since the coefficient in front of the (x)^2 term is Positive, this will be an upwards opening parabola in which the Vertex will be the MINIMUM Point on the parabola.

Therefore, the minimum output value will be y = -17 at the coordinate point (-2 , -17)

In other words, the minimum output value from any corresponding X-input value will be ——-> -17

Therefore, to ensure that the output value is (+)positive, we need to Shift the Parabola upwards along the Y Axis

If we Add + 17 outside the square’s term, this will shift the parabola upwards such that the vertex will now fall on the X-axis. That means we can still get an output of 0

So to ensure that we always have an output of greater than > 0

We must shift the parabola up by a little more than 17 units

And to do that we need to insert a value into n that is greater than > 17

n > 17

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