How do you graph y = 2x^ 2-13x-7y=2x213x7?

1 Answer
Feb 20, 2018

See below for a full explanation:

Explanation:

You can find the "zeroes (x-intercepts), y-intercepts, and the vertex."zeroes (x-intercepts), y-intercepts, and the vertex.

The standard form of a quadratic equation is

ax^2+bx+cax2+bx+c

with cc being the "y-intercept"y-intercept.

The "y-intercept"y-intercept of this equation is -77 as cc is -77.

To find the "zeroes"zeroes, plug 00 for yy. Use the quadratic formula.

(-b+-sqrt(b^2-4ac))/(2a)b±b24ac2a

Identify a,b,"and"a,b,and cc:

a=2a=2

b=-13b=13

c=-7c=7

Plug in:

(-(-13)+-sqrt((-13)^2-4*2*-7))/(2*2)=>(13)±(13)242722

(13+-sqrt(225))/4=>13±2254

(13+-15)/413±154

The zeroes are

7,-1/27,12

This was also factorable, as a discriminant (b^2-4acb24ac) that is a perfect square tells us that the equation is factorable (225225 is a perfect square).

You could factor and get

(2x+1)(x-7)=>(2x+1)(x7)

x=7,-1/2x=7,12

To get the vertex of an equation ax^2+bx+cax2+bx+c, use:

h=(-b)/(2a)h=b2a

k=c-(b^2)/4ak=cb24a

with (h,k)(h,k) being the vertex:

a=2a=2

b=-13b=13

c=-7c=7

h=(-(-13))/(2*2)=>13/4h=(13)22134

k=-7-((-13)^2)/(4*2)=>-56/8-169/8=>-225/8k=7(13)24256816982258

The vertex is (13/4,-225/8)(134,2258)

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Summary:

"y-intercept":(0,7)y-intercept:(0,7)

"x-intercepts": (7,0) andx-intercepts:(7,0)and (-1/2,0)(12,0)

"vertex":(13/4,-225/8)vertex:(134,2258)