Can someone help me with this pls??

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  a < -5  or  11 < a

Step-by-step explanation:

For there to be two points of intersection, there must be 2 real solutions to ...

  ax +9 = -2x^2 +3x +1

Rewriting to standard form, we have ...

  2x^2 +(a -3)x +8 = 0

The discriminant of this equation is ...

  B^2 -4AC = (a -3)^2 -4(2)(8) = a^2 -6a -55

We want this to be greater than 0. Factoring, we get ...

  (a -11)(a +5) > 0

This will be true for values of a that are less than -5 or greater than 11.

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The attached graph shows lines and their points of intersection for a=-6 and a=12. It is not hard to see that for slopes less steep than these values, the lines with a y-intercept of 9 may not intersect the parabola.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.