How do you use the important points to sketch the graph of #y=x^2-7x+12#?

Question

Julia Adams · Accepted Answer

graph{x^2-7x+12 [-12.17, 19.86, -0.56, 15.45]}

A quadratic function's key points are as follows: - vertex - y-intercept - x-intercept(s) Vertex: #(7/2,-1/4)# vertex: #y=x^2-7x+12# #y=(x^2-7x+49/4)+12-49/4# #y=(x-7/2)^2-1/4 -> y=a(x-h)^2+k# Complete the square to find #(h,k)#. See below. X-INTERCEPT(S): Factor the standard equation, setting each factor equal to zero. The values represent the x-coordinates for the x-intercepts: #y=x^2-7x+12#, #y=(x-4)(x-3)#, #x-4=0->x=4#, #x-3=0->x=3#. x-intercepts: (4,0), (3,0) Y-INTERCEPT: #y=x^2-7x+12 -> y=ax^2+bx+c# #a=1# #b=-7# #c=12# y-intercept: #(0,12)# We know that the c value in the equation is the y-coordinate of the y-intercept because the equation was provided in standard form. Graphing the parabola on an x and y plane is possible now that you have all the crucial points: graph{x^2-7x+12 [-12.17, 19.86, -0.56, 15.45]}

Eli Assouline · Answer

undefined To sketch the graph of $y = x^2 - 7x + 12$, you can follow these steps:

1. Find the vertex using the formula $x = \frac{-b}{2a}$.
2. Find the y-coordinate of the vertex by substituting the x-coordinate into the equation.
3. Determine the y-intercept by setting $x = 0$ and solving for y.
4. Find the x-intercepts by solving the equation $y = 0$ for x.
5. Plot the vertex, y-intercept, and x-intercepts on the coordinate plane.
6. Since the coefficient of $x^2$ is positive, the parabola opens upwards.
7. Sketch the parabola passing through these points symmetrically.