Graphing Quadratics Using Vertex and Intercepts

Question

Step 5 - Using the quadratic equation, find the "X" values or zeros of the graph. y = 4x^2 + 20x + 9

Step 5 -Using the quadratic equation, find the "X" values or zeros of the graph.   y = 5x^2 + 6x − 8

Answer 1

4x^2 +20x + 9

4(x^2 + 5x + (5/2)^2) + 9 -4(5/2)^2

= 4(x+5/2)^2 -16 = 0 (which is in vertex form. vertex = (-5/2, -16)= minimum point.

-5/2 is midway betwen the zeros, or their average.)

(x+5/2)^2 = 16/4 = 4

x+2.5 = +/-2

x = -0.5, -4.5 are the two zeros, values of x that make the original quadratic = 0

4(-1/2)^2 +20(-1/2) + 9= 1-10+9 = 10-10= 0

4(-9/2)^2+20(-9/2)+9 = 81-90+9 = 90-90=0

graphically, it's where the upward opening parabola intersects the x axis

-1/2 and -9/2 are the two zeros, roots, solutions or x intercepts

or graph the polynomial on a graphing calculator and look for where it crosses the x axis

5x^2 +6x -8

=5(x^2 + 6x/5 + (3/5)^2) -8 - 5(3/5)^2

=5(x+ 3/5)^2 - (40+9)/5 = 0

=5(x+3/5)^2 = 49/5

(x+3/5)^2 = 49/25

x+ 3/5 = +/-7/5

x = -3/5+/-7/5

x = -10/5, 4/5

x= -2, 0.8

5(-2)^2 +6(-2) -8

20-12-8

= 20-20 = 0

5(4/5)^2 +6(4/5)-8

= 16/5 +24/5 -8

= 40/5-8

= 8-8 = 0