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Use the standard form of a quadratic function, y=ax^2+bx+c to find a, b, and c.
y=- x^2+6x+16
We need at least three distinct pieces of information from a graph to find the quadratic that fits a parabola. In this case, we are asked to use the following three pieces of information.
Let's look at the graph and get these pieces of information from it.
We can start by drawing the axis of symmetry through the vertex.
From the axis of symmetry, we can write use the formula to solve for b in terms of a.
x= 3
Rearrange equation
LHS * 2a=RHS* 2a
.LHS /(-1).=.RHS /(-1).
Now, let's look at the y-intercept on the graph.
From the y-intercept we can conclude that c=16.
Let's choose one of the two marked x-intercepts.
We need to choose one zero, it does not matter which one. Let's choose the point (8,0) and substitute those values into our standard form of a quadratic.
y= 0, x= 8
Calculate power
b= -6a
Multiply
Subtract terms
c= 16
LHS-16a=RHS-16a
.LHS /-16.=.RHS /-16.
Now, we can use that and our equation we derived from the axis of symmetry to solve for b. cc b&=&-6 a &⇓& b&=&-6( -1) &⇓& b&=& 6 Since we now have a, b, and c, we can put those values into the standard quadratic form and obtain our function. y= ax^2+ bx+c ⇓ y = - x^2 + 6x + 16