Is it possible for the graph of a quadratic function to have equal x and y intercepts?Explain?

Answer 1

not sure quite what u mean, but y=x^2 has intercept at x=0, y=0

that any good?

Answer 2

the general quadratic equation is f(x) = a.x^2 + b.x + c

the question now is at what x is x=f(x)

if x=0 we have f(x)=c, y intercept

to have equal x-intercept, we must have that:
a.c^2+b.c+c = 0 or c(ac+b+1)=0

but we assume c not = 0 so ac+b+1 = 0 is the condition under which the equation will have equal intercepts

for eg: if a=1, c=1, b=-2, we have the condition satisfying....

so if f(x) = x^2 - 2x + 1 then we see that the x- & y-intercepts are equal, having value 1

Answer 3

Let the quadratic equation is
y =ax^2 + bx +c
To find the y intercept put x=0
Therefore y =c
Now find the x-intercept put y=0
then ax^2 +bx +c+0
Since x- intercept is equal to y- intercept therefore put x=c
ac^2 +bc +c=0
0r c(ac +b+1)=0 or c=0 and c=(-b-1)/a
It means that either c=0 or c = -(b+1)/a

Answer 4

Here is an example: Y=(x-3)^2 +3
This graph has a double root at x=3 (x-intercept) and a y intercept a 3 also.

Answer 5

Ha. College was a few years back for me. I really don't remember, but if you a TI graphing calculator put in the formula and graph it and it should tell what the x and y intercepts are.