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

Question

Eli Assouline · Accepted Answer

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

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

Genesis Allen · Answer

undefined To find the vertex of the quadratic function $f(x) = -7x^2 + 3x + 1$, you can use the formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex. Then, substitute this value back into the function to find the corresponding y-coordinate.

To find the intercepts, you can set $f(x) = 0$ and solve for $x$ to find the x-intercepts (or roots). Then, substitute these x-values back into the function to find the corresponding y-values for the y-intercepts.

Using these methods:

- The vertex of the function is at $\left(\frac{3}{14}, \frac{109}{28}$\).
- The x-intercepts are approximately $-0.159$ and $0.094$.
- The y-intercept is at $(0, 1)$.