Graph and discriminant

$\large y = x^2 + bx + c$ Curve passes through $(0,9)$ .This point must satisfy the equation. $\large c = 9$ $\large y = x^2 + bx + 9$ Because above equation has 2 real and equal roots, then D = 0 $\large \Rightarrow b^2 - 4\times 9 = 0$ $\large \Rightarrow b = |6|$

Since given curve touches $x$ axis so $y$ has some real root, say $m$ , of multiplicity $2$ . So $y$ can be written in the form of ${(x-m)}^{2}$ , so that $b$ turns out to be $-2m$ and $c$ turns out to be ${m}^{2}$ . From the second condition we can easily verify that $c=9$ . Thus ${m}^{2}=9$ and hence $|b|=6$

Vieta's Formulae tells us c/a = pq, where p and q are the roots. We have p = q, since there is only one x value making the function equal zero. Plugging in (0,9) we get c = 9 and pq = 9. Then p = q = 3. Again from Vieta, p + q = -b/a. 6 = -b. And|-6| = 6.

y = ax² + bx + c

y = x² + bx + c

Then, the value of a is 1.

(0,9) replacing the points x and y in the equation: y = x² + bx +c.
9 = 0² + b×0 + c, then, c = 9

Looking for the parable we can see that it has 2 real and equal roots, then, ∆ = 0

∆ = b² - 4ac

0 = b² - 4 × 1 × 9

b² = 4 × 9

b² = 36

b = |√36|

b = 6

Given that (0,9) is a point on the line, we can try to figure some things out by plug this point back into $y=x^{2}+bx+c$ .

$9=0^{2}+{0}\times{b}+c$

$c=9$

Now that we know what $c$ is, our function can now be written as $y=x^{2}+bx+c$ . In the problem, we are also given that this function touches the x-axis at some point, and when a function touches the x-axis, its y-value is equal to 0. If we plug 0 in for the y-variable in $y=x^{2}+bx+9$ , we get the quadratic $x^{2}+bx+9=0$ . What we need to find now is the b-value.

If you consider the fact that the b-value is the sum of the factors of 9, we the following possibilities:

$3+3=6$

$(-3)+(-3)=-6$

$1+9=10$

$(-1)+(-9)=-10$

However, the only possible b-value is $-6$ because $10$ and $-10$ 's is the sum of two different factors, and there can only be one factor because there is only one root (intersection with x-axis). The reason why $b=6$ is ignored is because if we plug 6 into our function, we get $x^{2}+6x+9=(x+3)(x+3)=0$ and solving for x would give us $-3$ , which is not the right x-value if the intersection is on the positive x-axis. Thus, $b=-6$ .

Finally, $\boxed{|-6|= 6}$ .

The answer is -6 because b^2=4*9 the b =+-6 but taking in consideration that it touches the positive x axes it must be -6 because (b-3)^2=0 gives you a positive answer not (b+3)^2=0 if try to drow it in a grapher you will know why is it -6 not 6

x=0, y=9; then c=9; now differentiate the given equation wrt x and then equate it to zero. Find the value of b in terms of x; ie b=2x; Now put this value & y=0 in the given equation; then |x|=3; Now b=2x=6; Hence the solution.