406 Graphs of Quadratic Functions
To find the x-values of the points where the line crosses the parabola, we have a new quadratic equa-
tion to solve:− 4 x^2 + 12 x− 9 =− 5We can put this into polynomial standard form by adding 5 to each side, getting− 4 x^2 + 12 x− 4 = 0The quadratic formula tells us that the roots of this equation arex= [−b± (b^2 − 4 ac)1/2] / (2a)
= {− 12 ± [12^2 − 4 × (−4)× (−4)]1/2} / [2 × (−4)]
= [− 12 ± (144 − 64)1/2] / (−8)
= [− 12 ± 80 1/2] / (−8)Figure 24-7 Approximate graph of y=− 4 x^2 + 12 x− 9,
where the independent variable is x and the
dependent variable is y. On both axes, each
increment represents 1/2 unit.xy(3/2,0)Each axis increment
is 1/2 unit(3/8,–5) where
x-value is
approximate(21/8,–5) where
x-value is
approximateLiney= –5