How do you find the vertex and intercepts for f(x) = -7x^2 + 3x + 1f(x)=−7x2+3x+1?

1 Answer
Nallasivam V
Jul 27, 2018

Vertex (3/14,37/28)(314,3728)
y-intercept (0,1)(0,1)

x-intercept ((sqrt37+3)/14,0); ((-sqrt37+3)/14, 0)(√37+314,0);(−√37+314,0)

Explanation:

Give -

f(x) =-7x^2+3x+1f(x)=−7x2+3x+1

y=-7x^2+3x+1y=−7x2+3x+1

Vertex

x=(-b)/(2a)=(-3)/(2 xx -7)=(-3)/(-14)=3/14x=−b2a=−32×−7=−3−14=314

At x=3/14; y=-7(3/14)^2+3(3/14)+1x=314;y=−7(314)2+3(314)+1

y=-7(9/196)+9/14+1=(-63+126+196)/196=259/196=37/28y=−7(9196)+914+1=−63+126+196196=259196=3728

(3/14,37/28)(314,3728)

y-intercept

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

y-intercept (0,1)(0,1)

x-intercept

Aty=0; -7x^2+3x+1=0y=0;−7x2+3x+1=0

-7x^2+3x=-1−7x2+3x=−1

x^2-3/7x+9/196=-1/(-7)+9/196=(28+9)/196=37/196x2−37x+9196=−1−7+9196=28+9196=37196

(x-3/14)^2=+-sqrt(37/196)=+-sqrt37/14(x−314)2=±√37196=±√3714

x=sqrt37/14+3/14=(sqrt37+3)/14x=√3714+314=√37+314
x=-sqrt37/14+3/14=(-sqrt37+3)/14x=−√3714+314=−√37+314

x-intercept ((sqrt37+3)/14,0); ((-sqrt37+3)/14, 0)(√37+314,0);(−√37+314,0)

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