How do I find the x-intercepts for the graph of #y=x^2-3x-4#? Precalculus Geometry of a Parabola Graphing Parabolas 1 Answer Ernest Z. Jul 9, 2015 You set #y = 0# and solve for #x#. Explanation: #y = x^2–3x-4 = 0# #(x+1)(x-4) = 0# #x+1 = 0 or x-4 = 0# #x = -1# or #x = 4# The #x#-intercepts are at (#-1,0#) and (#4,0#). Answer link Related questions What is the graph of the parabola represented by #y=2x^2-8x+9#? What is the parent graph of a parabola? How do I find the axis of the graph of the function #f(x)=4x^2+8x+7#? What is the graph of #f(x)=-2x^2+7x+4#? What is the x-intercept of the graph of #y=x^2-4x+4#? What is the vertex of the graph of #y=x^2+3x-4#? What is the vertex of the graph of #y=2(x-3)^2-7#? What is the vertex of the graph of #y=-(x-2)^2+3#? What is the axis of symmetry of the graph of #y=-(x+3)^2-6#? How do I graph #y=x^2-2x-3#? See all questions in Graphing Parabolas Impact of this question 5392 views around the world You can reuse this answer Creative Commons License