The equation of the quadratic function in standard form as required in the task content is; f(x) = -x² + 12x - 43.
<h3>Standard form equation of a quadratic function.</h3>
It follows from the task content that the standard form equation of the quadratic function is to.be determined.
Since the standard form equation can be derived from the vertex form equation as follows;
f(x) = a (x - h)² + k
f(x) = a (x - 6)² - 7
Hence, to find the value of a, Substitute x = 8 and f(x) = -11 so that we have;
-11 = a (8 - 6)² - 7
-11 = 4a - 7
4a = -4
a = -1.
Hence, the equation in vertex form is; f(x) = -1 (x -6)² - 7 and when expressed in standard form we have;
f(x) = -1(x² - 12x + 36) - 7
f(x) = -x² + 12x - 43
Therefore, the required equation in standard form is; f(x) = -x² + 12x - 43.
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Since both are equivalent to y, the equations must be equivalent.
x^2-x-3= -3x+5
x^2+2x-8=0
(x+4)(x-2)=0
x=-4, x=2
Plug the values of x in to either equation
y=-3(-4)+5
y= 12+5
y=17
y= -3(2)+5
y=-6+5
y=-1
Final answer: (-4,17) and (2,-1)
FD/CA = EF/BC
x/(8.5 mm) = (12 mm)/(4 mm)
x = (8.5 mm)*(12/4)
x = 25.5 mm . . . . . . . . . . . the 3rd selection
I personally use y2-y1 over x2-x1 because it's simple. Plot two y values and two x values. The first ones you plotted are x1 and y1, the others are x2 y2. Now subtract y2-y1 and x2-x1. Afterwards, divide the y value by the x value, that will give you slope.
Answer:
The 2nd term is 3y
Step-by-step explanation:
2x + 3y
__ __
1 2
term term
