Need help with this have no idea how to do it
1 answer:
Answer:
Step-by-step explanation:
x- intercept, vertex (v) and axis of symmetry, parabola form:x²+bx+c
vertex(h,k)
<h2>1- f(x)=(x+3)(x-3) change into parabola form </h2>f(x)=x²-9 a=1,b=0 ,c=-9
h=-b/2a=0
k=f(0)=-9
vertex(0,-9) , x intercept is when f(x)=0
(x-3)(x+3)=0 either x-3=0⇒ x=3 or x+3=0 then x=-3
x=3, x=-3 (-3,0) and (3,0)
the x of symmetry is the h of the vertex=0
<h2>2-g(x)=(x+1)(x-3)</h2>g(x)=x²-2x-3 a=1, b=-2,c=-3
h=-b/2a⇒-(-2)/2(1)⇒h=1
k=f(1)=1²-2(1)-3⇒k=-4
v(1,-4)
x of symmetry=h=1
(x+1)(x-3)=0
x+1=0⇒x=-1 (-1,0)
x-3=0 ⇒x=3 (3,0)
x intercept :-1,3
<h2>3-y=-x(x+6) ⇒y=-x²-6x a=-1,b=-6, c=0</h2>vertex(-3,9)
x intercept:(0,0) and (-6,0)
axis of symmetry =-3
<h2>4-g(x)=2(x-5)(x-1) ⇒ 2x²-12x+10 a=2, b=-12, c=10</h2>vertex(3,-8)
axis of symmetry=3
x intercept : (5,0), (1,0)
<h2>5) -4x(x+1)⇒-4x²-4x a=-4,b=-4</h2>vertex(-1/2,1)
x of symmetry=-1/2
x intercept : (0,0)(-1,0)
<h2>6- f(x)=-2(x-3)² ⇒-2x²+12x-18</h2>vertex(3,0)
x intercept (3,0)
axis of symmetry = 3
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We could use the formula, derive the formula, or just work it out for this case. Let's do the latter.
The distance of a point to a line is the length of the perpendicular from the line to the point.
So we need the perpendicular to 5x-4y=10 through (-1,3). To get the perpendicular family we swap x and y coefficients, negating one. We get the constant straightforwardly from the point we're going through:
4x + 5y = 4(-1)+5(3) = 11
Those lines meet at the foot of the perpendicular, which is what we're after.
4x + 5y = 11
5 x - 4y = 10
We eliminate y by multiplying the first by four, the second by five and adding.
16x + 20y = 44
25x - 20y = 50
41x = 94
x = 94/41
y = (11 - 4x)/5 = 15/41
We want the distance from (-1,3) to (94/41,15/41)