How do you find the vertex and the intercepts for #y=-x^2+x+12#?

Question

Hazel Anglin · Accepted Answer

Vertex: #(1/2,49/4)#
y-intercept: #12#
x-intercepts: #4# and #(-3)# The vertex form of a parabola is #color(white)("XXX")y=color(m)(x-color(blue)(a))+color(red)(b)# with vertex at #(color(blue)(a),color(red)(b))# Converting the given #y=-x^2+x+12# into vertex form: #color(white)("XXX")y=color(green)((-1))(x^2-x+(1/2)^2)+12 + (1/2)^2# #color(white)("XXX")y=color(green)((-1))(x-color(blue)(1/2))^2+color(red)(49/4)# giving us the vertex at #(color(blue)(1/2,)color(red)(49/4))# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The y-intercept is the value of #y# when #color(magenta)(x=0)# #y=-x^2+x+12# when #color(magenta)(x=0)# #rarr y=-color(magenta)(0)^2+color(magenta)(0)+12=12# So the y-intercept is #0# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The x-intercepts are the values of #x# when #color(brown)(y=0)# #y=-x^2+x+12# when #color(brown)(y=0)# #rarr -x^2+x+12=color(brown)(0) color(white)("XX")orcolor(white)("XX")x^2-x-12=0# This can be factored as #color(white)("XXX")(x-4)(x+3)=0# #rarr x=4color(white)("XX")orcolor(white)("XX")x=-3# So the x-intercepts are #{4,-3}# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Here is the graph of this relation for verification purposes: graph{-x^2+x+12 [-10.29, 18.19, -1.36, 12.9]}

Jace Anderson · Answer

undefined To find the vertex of the parabola described by the equation y = -x^2 + x + 12, you can use the formula for the x-coordinate of the vertex, which is given by x = -b/(2a) where a and b are the coefficients of the quadratic equation in the form ax^2 + bx + c. In this equation, a = -1 and b = 1.

To find the y-coordinate of the vertex, substitute the x-coordinate obtained into the original equation.

To find the x-intercepts, set y = 0 and solve the resulting quadratic equation for x.

To find the y-intercept, set x = 0 and solve the equation for y.

The vertex coordinates are (-b/(2a), f(-b/(2a))), where f(x) is the function represented by the equation.

In this case, the x-coordinate of the vertex is x = 1/2, and substituting x = 1/2 into the original equation gives y = 47/4.

The x-intercepts can be found by solving -x^2 + x + 12 = 0.

The y-intercept is found by substituting x = 0 into the equation.

So, the vertex of the parabola is (1/2, 47/4), the x-intercepts are approximately (-3, 0) and (4, 0), and the y-intercept is (0, 12).