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What is the graph of #f(x)=-2x^2+7x+4#?

Answer 1

Elijah Allen

check the explanation below.

#y=-2x^2+7x+4#

After taking -2 as a common factor between the first two terms, finish the square.

#y=-2(x^2-7/2x)+4#

#y=-2((x-7/4)^2-(7/4)^2)+4#

#y=-2(x-7/4)^2+10.125#

it's vertex is #(7/4,10.125)#

supplementary points

It's intersection with the #x-"axis"#

and opened downwards since the coefficient of #x^2# is negative

#y=0rarr x=-0.5 or x=4#

graph{y=-21.56, 13.76, -1.42, 11.24]}

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Answer 2

Sadie Ackerman

The graph of ( f(x) = -2x^2 + 7x + 4 ) is a downward-opening parabola because the coefficient of ( x^2 ) is negative. The vertex of the parabola can be found using the formula ( x = \frac{-b}{2a} ), where ( a = -2 ) and ( b = 7 ). Once you find the x-coordinate of the vertex, you can substitute it into the function to find the y-coordinate. Additionally, you can find the x-intercepts by setting ( f(x) = 0 ) and solving the resulting quadratic equation. The y-intercept can be found by evaluating the function when ( x = 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.