How do you write the quadratic function in intercept form given x intercepts 3,9 and point (14,77)?

Question

Isaiah Anthony · Accepted Answer

#y=7/5(x-3)(x-9)# #"given x-intercepts (zeros) say "x=c,x=d# #"then the factors are "(x-c)" and "(x-d)# #"and y is the product of the factors"# #"here "x=3" and "x=9# #rArr(x-3)" and "(x-9)" are the factors"# #rArry=a(x-3)(x-9)larrcolor(blue)"a is a multiplier"# #"to find a substitute "(14,77)" into the equation"# #77=a(11)(5)# #rArra=77/55=7/5# #rArry=7/5(x-3)(x-9)larrcolor(red)"in intercept form"#

Jace Anderson · Answer

undefined To write the quadratic function in intercept form given x-intercepts 3 and 9, and a point (14, 77), we first recognize that the x-intercepts provide the roots of the quadratic equation, while the point (14, 77) gives us another point through which the graph passes.

Using the x-intercepts, we can write the quadratic function as $ f(x) = a(x - 3)(x - 9) $, where $ a $ is the leading coefficient.

Substitute the point (14, 77) into the function to find the value of $ a $:
$$ 77 = a(14 - 3)(14 - 9) $$
$$ 77 = a(11)(5) $$
$$ 77 = 55a $$
$$ a = \frac{77}{55} $$

So the quadratic function in intercept form is:
$$ f(x) = \frac{77}{55}(x - 3)(x - 9) $$