Find b and c so that y=-1x^2+bx+c has vertex (-5,-4). Please help ASAP I'm having a really hard time on this question!!!
do you think you can help me check my question
i think i made a typo b=10
not -10
because of the negative value of a
that shouldn't be a problem right
the ideal is the same,so you do the correction
You want y=-1x^2+bx+c with vertex at (-5,-4)
The vertex at (-5,-4) means that
y = a(x+5)^2 - 4
So now you have
ax^2 + 10ax + 21 = -x^2+bx+c
(a,b,c) = (-1,10,21)
To find the values of b and c in the equation y = -1x^2 + bx + c, given that the vertex is (-5,-4), we'll need to use the concept of vertex form.
The vertex form of a quadratic equation is given by y = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex.
In this case, we have the vertex (-5, -4), so the equation can be rewritten as y = a(x + 5)^2 - 4.
Comparing this with the original equation y = -1x^2 + bx + c, we can see that a = -1.
Now, we can substitute the values of a (-1) and the coordinates of the vertex (-5, -4) into the vertex form equation:
-4 = -1(-5 + 5)^2 - 4
-4 = -1(0)^2 - 4
-4 = 0 - 4
-4 = -4
Since the equation holds true, we can conclude that b = 0 and c = -4.
Therefore, the values of b and c that satisfy the given equation are b = 0 and c = -4.
parabola equation y=ax²+bx+c
the x- coordinate is -b/2a
so for
y=-x²+bx+c
ax²+bx+c=-x²+bx+c
a=-1
b/2(1)=-5
b=-10
hence you have the equation to be
y=-x²-10x+c
x²+10x-c=-y
(2)²+10(2)-c=4
4+20-c=4
24-c=4
-c=4-24
c=20