How do you use the important points to sketch the graph of #y=x^2-4x+7#?

Question

Genesis Allen · Accepted Answer

The graph intersects the y-axis at #(0, 7)# and the vertex is at #(2, 3)#
and the term #y# carries a + sign, therefore the parabola opens upward.

Vertex calculation using the "Completing the square method" The vertex form #y=x^2-4x+7#, #y=x^2-4x+4-4+7#, #y=(x-2)^2+3#, #y-3=(x-2)^2#, and #(x-2)^2=y-3# displays the vertex at #(2, 3)#. Finding the intercepts in the formula #y=x^2-4x+7# when #x=0#, #y=0^2-4*0+7#, and #y=7# A point on the y-axis and parabola is #(0, 7)#. #y=x^2-4x+7# graph{y=x^2-4x+7[-20,20,-10,10]} is the graph displayed. May God bless you all. I hope this explanation helps.

Eva Adamson · Answer

undefined To sketch the graph of $ y = x^2 - 4x + 7 $ using important points:

1. Identify the vertex of the parabola using the formula $ x = \frac{-b}{2a} $, where $ a = 1 $ and $ b = -4 $. Calculate $ x $ coordinate, then substitute it into the equation to find the corresponding $ y $ coordinate.

2. Find the $ y $-intercept by substituting $ x = 0 $ into the equation.

3. Locate the $ x $-intercepts by setting $ y = 0 $ and solving for $ x $ using the quadratic formula.

4. Use the axis of symmetry (the $ x $-coordinate of the vertex) to find additional points symmetrically placed on either side.

5. Plot the points and sketch the parabolic curve passing through them.