I need this graphed y=4x^2+3x+1

Question

we are learning graphing quadratic functions

Answer 1

Find some key points.

What's the y intercept?

Are there any zeros (x-intercepts, or 'zeros')? Find them by factoring, if possible. (It isn't)

Since discriminant < 0, there are no zeros.

But there is an axis of symmetry, the line x = -b/(2a). This takes you to the vertex, so that's a second plottable point.

 

To get a few other points, substitute various values of x...

Answer 2

It's a quadratic so you know it's a parabola!
Find the turning point ( or vertex), axis of symmetry, intercepts...

 

 

y = 4x²+3x+1

a=4 and positive, so the graph goes upwards!

it's also been narrowed by 4 compared to the basic y=x² graph

b=3

c=1

 

h is how far the curve shifted left of right, it's the axis of symmetry. Recall that the tuning point is always on the axis of symmetry. So when we find this value, we've found the x value of the turning point.

h = -b/2a

h= -3/(2)(4)

h = -3/8

since this is negative, the graph is shifted to the left by 3/8

pop the value of h into the original equation to get it's y coordinate.

y = 4(-3/8)²+3(-3/8) + 1

y = 4 (9/64) + -9/8 + 1

y = 9/16 - 9/8 + 1

get a common denominator of 16

y = (9 - 18 + 16)/16

y = 7/16

 

our turning point is 
(-3/8, 7/16)

 

 

 

set x=0 and y will be

y = 1

 

and you've enough info to graph!