# How do you find the vertex and intercepts for y=x^2+10x+21?

Dec 9, 2015

vertex $\to \left(x , y\right) \to \left(- 5 , - 4\right)$

${y}_{\text{intercept}} = 21$

x_("intercept")=-3 " and " -7

#### Explanation:

Given: $\textcolor{b l u e}{y = {x}^{2} + 10 x + 21}$

$\textcolor{b r o w n}{\text{To find the vertex}}$

Consider the +10 from +10x

$\textcolor{b r o w n}{{x}_{\text{vertex}}} = \left(- \frac{1}{2}\right) \times \left(+ 10\right) \textcolor{b r o w n}{= - 5}$

Substitute the found value of ${x}_{\text{vertex}}$ in the original equation to find ${y}_{\text{vertex}}$

${y}_{\text{vertex}} = {\left(- 5\right)}^{2} + 10 \left(- 5\right) + 21$

color(brown)(y_("vertex")=-4

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Each intercept is when the plotted line crosses one of the axis. That is: when $x = 0$ the plot crosses ${y}_{\text{intercept}}$ and when $y = 0$ the plot crosses ${x}_{\text{intercept}}$

$\textcolor{b r o w n}{\text{So to find "y_("intercept") " write the equation as:}}$

${y}_{\text{intercept}} = {0}^{2} + 10 \left(0\right) + 21$

$\textcolor{b r o w n}{{y}_{\text{intercept}} = + 21}$

You will observe from this that it is the value of the constant.
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$\textcolor{b r o w n}{\text{So to find "x_("intercept") " write the equation as:}}$

$0 = {x}^{2} + 10 x + 21$

I notice that $3 \times 7 = 21$ and $3 + 7 = 10$

so we can factorise giving:

$\textcolor{b l u e}{0 = \left(x + 3\right) \left(x + 7\right)} \textcolor{g r e e n}{\to {x}^{2} + 7 x + 3 x + 21} = {x}^{2} + 10 x + 21$

So we have to make each of the bracketed parts 0 to give $y = 0$

So for $\left(x + 3\right) = 0$ the value of $x = - 3$
and for $\left(x + 7\right) = 0$ the value of $x = - 7$

So color(brown)(x_("intercept") = -7 " and " -3)