What are the important points needed to graph y = 3x^2 + 8x - 6y=3x2+8x−6?

1 Answer
Nallasivam V
Oct 28, 2015

Its vertex is ((-4)/3, (-2)/3)(−43,−23)
Since the co-efficient of x^2x2 is positive, the curve is open upwards.

It has a minimum at ((-4)/3, (-2)/3)(−43,−23)
Its y- intercept is -6−6

Explanation:

Given-

y=3x^2+8x-6y=3x2+8x−6

We have to find the vertex

x=(-b)/(2a)=(-8)/(2 xx 3)=(-8)/6=(-4)/3x=−b2a=−82×3=−86=−43

At x=(-4)/3x=−43;

y=3((-4)/3)^2+8((-4)/3)-6y=3(−43)2+8(−43)−6
y=3((16)/9)-32/3-6y=3(169)−323−6
y=48/3-32/3-6=(-2)/3y=483−323−6=−23

Its vertex is ((-4)/3, (-2)/3)(−43,−23)

Take two points on either side of x=(-4)/3x=−43

Find the y values. Plot the points. Join them with a smooth curve.

Since the co-efficient of x^2x2 is positive, the curve is open upwards.

It has a minimum at ((-4)/3, (-2)/3)(−43,−23)

Its y- intercept is -6−6

Since the co-efficient of x^2x2 is 3, the curve is a narrow one.

graph{3x^2+8x-6 [-25.65, 25.65, -12.83, 12.82]}

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