How do you find the vertex and intercepts for # f(x) = -7x^2 + 3x + 1#?

1 Answer
Nallasivam V
Jul 27, 2018

Vertex #(3/14,37/28)#
y-intercept #(0,1)#

x-intercept #((sqrt37+3)/14,0); ((-sqrt37+3)/14, 0)#

Explanation:

Give -

#f(x) =-7x^2+3x+1#

#y=-7x^2+3x+1#

Vertex

#x=(-b)/(2a)=(-3)/(2 xx -7)=(-3)/(-14)=3/14#

At #x=3/14; y=-7(3/14)^2+3(3/14)+1#

#y=-7(9/196)+9/14+1=(-63+126+196)/196=259/196=37/28#

#(3/14,37/28)#

y-intercept

At #x=0; y=-7(0)+3(0)+1=1#

y-intercept #(0,1)#

x-intercept

At#y=0; -7x^2+3x+1=0#

#-7x^2+3x=-1#

#x^2-3/7x+9/196=-1/(-7)+9/196=(28+9)/196=37/196#

#(x-3/14)^2=+-sqrt(37/196)=+-sqrt37/14#

#x=sqrt37/14+3/14=(sqrt37+3)/14#
#x=-sqrt37/14+3/14=(-sqrt37+3)/14#

x-intercept #((sqrt37+3)/14,0); ((-sqrt37+3)/14, 0)#

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
1980 views around the world
You can reuse this answer
Creative Commons License