Remember that the vertex form of a parabola or quadratic equation is:

y=a(x-h)^2+k, where (h,k) is the "vertex" which is the maximum or minimum point of the parabola (and a is half the acceleration of the of the function, but that is maybe too much :P)

In this case we are given that the vertex is (1,1) so we have:

y=a(x-1)^2+1, and then we are told that there is a point (0,-3) so we can say:

-3=a(0-1)^2+1

-3=a+1

-4=a so our complete equation in vertex form is:

y=-4(x-1)^2+1

Now you wish to know where the x-intercepts are. x-intercepts are when the graph touches the x-axis, ie, when y=0 so

0=-4(x-1)^2+1 add 4(x-1)^2 to both sides

4(x-1)^2=1 divide both sides by 4

(x-1)^2=1/4 take the square root of both sides

x-1=±√(1/4) which is equal to

x-1=±1/2 add 1 to both sides

x=1±1/2

So x=0.5 and 1.5, thus the x-intercept points are:

(0.5, 0) and (1.5, 0) or if you like fractions:

(1/2, 0) and (3/2, 0) :P

4 0

3 years ago

Need help urgently please

kakasveta [241]

Answer:

x = -12

y = 3

Step-by-step explanation:

$1/2x + 6(x+15) = 12\\1/2x + 6x + 90 = 12\\6\frac{1}{2} x + 90 = 12\\6\frac{1}{2} x= -78\\x = -12$