Question

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

Answer 1

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)`