Question

Find the x and y intercepts.
f(x)=4^2+20x+25
Y=
x=

Answer 1

the y intercept occurs when x = 0

so substitute x = 0... and find y

x intercepts occur when y = 0

to let f(x) = 0 and solve for x

is your equation

\[f(x) = 4x^2 + 20x + 25 \]

if it is then you have

\[4x^2 + 20x + 25 = 0\]

this is a perfect square as 4x^2 = (2x)^2 and 25 = (5)^2

so it can be factorised to

\[(x + 5)^2 = 0\]

just solve for x to get the x-intercept

Answer 2

x=-5?

Answer 3

yes thats the x-intercept

Answer 4

How do I get y?

Answer 5

let x = 0

so y = 4(0)^2 + 20(0) + 25

Answer 6

your y intercept is y = 25

Answer 7

Is y 25??? I'm confused

Answer 8

@campbell_st I think you got the x- intercept wrong

4x^2 + 20x + 25 = 0

(divide by 4) x^2 + 5x + 25/4 = 0

(group the x's) x^2 +5x = -25/4

(half x-term coefficient and square)
5 = 5/2 = 25/4

(add to both sides)
x^2 + 5x + 25/4 = -25/4 + 25/4

x^2 + 5x + 25/4 = 0

\[\left( x+\frac{5}{2} \right)^2 = 0\]

(square both sides and solve for x)

\[\Large x = -\frac{5}{2} \pm \sqrt{0} \longrightarrow x _{1,2} = -\frac{5}{2}\]

So the x-intercept should be -5/2

Answer 9

oops (2x +5)^2 =0

x = -5/2 and y = 25

Answer 10

Thanks!!! ;-)