so here is the question

graph the function. state the domain and range of the function.

F(x)=-2x^2-3x+2

i try and use the axis of symmetry (-b/2a) and then plug that into the original formula to get the vertex.
i keep getting F(-3/2)=2

but then when i try and plug in numbers into the formula there scattered everywhere

F(-3/2)=2 is the vertex in the graph but i cant find the other points to show where the parabola is going

i tried -1/2, 0, -2 and got
(-1/2,3)
(0,-3)
(-2,0)

its supposed to be a upside down parabola with a domain of all real number and a range of {y|y<_ 25/8}

<_ = less than or equal to.

Last edited:

arildno
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so here is the question

graph the function. state the domain and range of the function.

F(x)=-2x^2-3x+2

i try and use the axis of symmetry (-b/2a) and then plug that into the formula for the vertex. i keep getting F(-3/2)=2
Which is correct.

Your axis of symmetry is, however, x=-3/4.

The maximum value of F is therefore:
$$F(-\frac{3}{4})=-2(-\frac{3}{4})^{2}-3*\frac{-3}{4}+2=\frac{-18}{16}+\frac{9}{4}+2=\frac{50}{16}=\frac{25}{8}$$

ughhhh i put -1 for a not -2
thanks.