The vertex of the graph of a quadratic function is in the second quadrant. The related equation has no real solutions. Which statement is true?
a) The graph opens up.
b) The graph opens down.
c) The y-intercept is 0.
d) The axis of symmetry is x = 0.

a) The graph opens up.

Step-by-step explanation:

The vertex is the minimum/maximum of a parabola.

We know that the vertex is in the second quadrant (so it is in the quadrant of positive y values and negative x values)

We also know that it has no real roots, so the graph never touches the x-axis, and knowing that the vertex is above the x-axis, then the graph must open upwards.

Then the correct option is:

a) The graph opens up.

i think this could be a possible answer (1 / 84) -1

good luck! : d

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.