How do you find the vertex, axis of symmetry and intercepts of #f(x)=x^2+3x+7#?

Question

Nathan Axel · Accepted Answer

Vertex #(-3/2, 19/4)#

Axis of symmetry #x=-3/2# Given - #y=x^2+3x+7# coordinate of the vertex in x #x=(-b)/2a=(-3)/(2 xx 1)=-3/2# y coordinate of the vertex At #x=-3/2# #y=(-3/2)^2+3(-3/2)+7# #y=9/4-9/2+7=(9-18+28)/4=19/4# Vertex #(-3/2, 19/4)# Axis of symmetry #x=-3/2#

Eva Abramson · Answer

undefined To find the vertex of the quadratic function $ f(x) = x^2 + 3x + 7 $, use the formula $ x = \frac{-b}{2a} $, where $ a $ is the coefficient of $ x^2 $, and $ b $ is the coefficient of $ x $. Thus, $ x = \frac{-3}{2} $. Substitute this value of $ x $ into the function to find the corresponding $ y $-coordinate. This gives the vertex $ ( -\frac{3}{2}, \frac{17}{4}) $.

The axis of symmetry is the vertical line that passes through the vertex. For this function, the axis of symmetry is $ x = -\frac{3}{2} $.

To find the intercepts, set $ f(x) $ equal to zero to find the $ x $-intercepts (or roots). Use the quadratic formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $. Substitute the values of $ a $, $ b $, and $ c $ from the function. This gives the discriminant $ b^2 - 4ac = 3^2 - 4(1)(7) = -19 $, which is negative, indicating that the function does not have real roots. Thus, it has no $ x $-intercepts.

To find the $ y $-intercept, set $ x = 0 $ in the function $ f(x) $. This gives $ f(0) = 7 $. So, the $ y $-intercept is $ (0, 7) $.